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4 X 2 X 30

$\exponential{ten}{2} - four x - 30 = 0 $

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x^{2}-4x-thirty=0

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and i when it is subtraction.

ten=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\left(-thirty\right)}}{two}

This equation is in standard course: ax^{two}+bx+c=0. Substitute 1 for a, -four for b, and -xxx for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

ten=\frac{-\left(-4\correct)±\sqrt{16-4\left(-xxx\right)}}{ii}

Square -4.

x=\frac{-\left(-four\right)±\sqrt{16+120}}{2}

Multiply -4 times -xxx.

x=\frac{-\left(-4\right)±\sqrt{136}}{two}

Add together 16 to 120.

x=\frac{-\left(-4\right)±2\sqrt{34}}{2}

Take the square root of 136.

x=\frac{4±2\sqrt{34}}{2}

The opposite of -four is 4.

x=\frac{2\sqrt{34}+4}{2}

Now solve the equation x=\frac{4±2\sqrt{34}}{2} when ± is plus. Add 4 to 2\sqrt{34}.

10=\sqrt{34}+ii

Carve up 4+2\sqrt{34} by ii.

x=\frac{iv-2\sqrt{34}}{2}

At present solve the equation x=\frac{4±2\sqrt{34}}{2} when ± is minus. Subtract 2\sqrt{34} from 4.

10=2-\sqrt{34}

Split iv-2\sqrt{34} by two.

ten=\sqrt{34}+ii 10=ii-\sqrt{34}

The equation is now solved.

x^{two}-4x-30=0

Quadratic equations such every bit this one tin be solved by completing the square. In club to complete the square, the equation must first exist in the course ten^{2}+bx=c.

x^{2}-4x-xxx-\left(-30\right)=-\left(-thirty\correct)

Add thirty to both sides of the equation.

ten^{2}-4x=-\left(-xxx\right)

Subtracting -30 from itself leaves 0.

10^{two}-4x=30

Subtract -30 from 0.

ten^{2}-4x+\left(-2\right)^{2}=30+\left(-ii\correct)^{two}

Divide -4, the coefficient of the ten term, past 2 to get -2. Then add the foursquare of -ii to both sides of the equation. This step makes the left hand side of the equation a perfect foursquare.

x^{ii}-4x+4=30+4

Foursquare -two.

x^{ii}-4x+4=34

Add together 30 to 4.

\left(ten-2\correct)^{2}=34

Cistron x^{2}-4x+4. In general, when x^{2}+bx+c is a perfect foursquare, it can always exist factored equally \left(x+\frac{b}{2}\correct)^{2}.

\sqrt{\left(ten-2\right)^{ii}}=\sqrt{34}

Take the square root of both sides of the equation.

x-two=\sqrt{34} x-2=-\sqrt{34}

Simplify.

x=\sqrt{34}+2 x=2-\sqrt{34}

Add 2 to both sides of the equation.

ten ^ ii -4x -30 = 0

Quadratic equations such every bit this one tin be solved by a new direct factoring method that does not require guess work. To use the straight factoring method, the equation must be in the grade x^2+Bx+C=0.

r + s = 4 rs = -30

Let r and south be the factors for the quadratic equation such that x^two+Bx+C=(x−r)(ten−south) where sum of factors (r+due south)=−B and the product of factors rs = C

r = 2 - u s = 2 + u

Two numbers r and s sum up to four exactly when the boilerplate of the ii numbers is \frac{1}{ii}*4 = 2. You lot tin too run into that the midpoint of r and s corresponds to the axis of symmetry of the parabola represented by the quadratic equation y=x^2+Bx+C. The values of r and s are equidistant from the eye past an unknown quantity u. Express r and s with respect to variable u. <div way='padding: 8px'><img src='https://opalmath.azureedge.cyberspace/customsolver/quadraticgraph.png' way='width: 100%;max-width: 700px' /></div>

(2 - u) (2 + u) = -30

To solve for unknown quantity u, substitute these in the product equation rs = -30

4 - u^two = -30

Simplify past expanding (a -b) (a + b) = a^2 – b^ii

-u^2 = -30-4 = -34

Simplify the expression by subtracting 4 on both sides

u^two = 34 u = \pm\sqrt{34} = \pm \sqrt{34}

Simplify the expression by multiplying -1 on both sides and have the square root to obtain the value of unknown variable u

r =two - \sqrt{34} = -3.831 s = 2 + \sqrt{34} = seven.831

The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.

4 X 2 X 30,

Source: https://mathsolver.microsoft.com/en/solve-problem/x%20%5E%20%7B%202%20%7D%20-%204%20x%20-%2030%20%3D%200

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