Solve 8x+40=4x^2-56

$$8x+40=4 { x }^{ 2 } -56$$

Solve for x

$$x=-4$$

$$x=6$$

Steps Using Factoring By Grouping

Subtract 4x^{2} from both sides.

$$8x+40-4x^{2}=-56$$

Add 56 to both sides.

$$8x+40-4x^{2}+56=0$$

Add 40 and 56 to get 96.

$$8x+96-4x^{2}=0$$

Divide both sides by 4.

$$2x+24-x^{2}=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$-x^{2}+2x+24=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx+24. To find a and b, set up a system to be solved.

$$a+b=2 ab=-24=-24$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -24.

$$-1,24 -2,12 -3,8 -4,6$$

Calculate the sum for each pair.

$$-1+24=23 -2+12=10 -3+8=5 -4+6=2$$

The solution is the pair that gives sum 2.

$$a=6 b=-4$$

Rewrite -x^{2}+2x+24 as \left(-x^{2}+6x\right)+\left(-4x+24\right).

$$\left(-x^{2}+6x\right)+\left(-4x+24\right)$$

Factor out -x in the first and -4 in the second group.

$$-x\left(x-6\right)-4\left(x-6\right)$$

Factor out common term x-6 by using distributive property.

$$\left(x-6\right)\left(-x-4\right)$$

To find equation solutions, solve x-6=0 and -x-4=0.

$$x=6 x=-4$$

Steps Using the Quadratic Formula

Subtract 4x^{2} from both sides.

$$8x+40-4x^{2}=-56$$

Add 56 to both sides.

$$8x+40-4x^{2}+56=0$$

Add 40 and 56 to get 96.

$$8x+96-4x^{2}=0$$

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 one when it is subtraction.

$$-4x^{2}+8x+96=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -4 for a, 8 for b, and 96 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-8±\sqrt{8^{2}-4\left(-4\right)\times 96}}{2\left(-4\right)}$$

Square 8.

$$x=\frac{-8±\sqrt{64-4\left(-4\right)\times 96}}{2\left(-4\right)}$$

Multiply -4 times -4.

$$x=\frac{-8±\sqrt{64+16\times 96}}{2\left(-4\right)}$$

Multiply 16 times 96.

$$x=\frac{-8±\sqrt{64+1536}}{2\left(-4\right)}$$

Add 64 to 1536.

$$x=\frac{-8±\sqrt{1600}}{2\left(-4\right)}$$

Take the square root of 1600.

$$x=\frac{-8±40}{2\left(-4\right)}$$

Multiply 2 times -4.

$$x=\frac{-8±40}{-8}$$

Now solve the equation x=\frac{-8±40}{-8} when ± is plus. Add -8 to 40.

$$x=\frac{32}{-8}$$

Divide 32 by -8.

$$x=-4$$

Now solve the equation x=\frac{-8±40}{-8} when ± is minus. Subtract 40 from -8.

$$x=-\frac{48}{-8}$$

Divide -48 by -8.

$$x=6$$

The equation is now solved.

$$x=-4 x=6$$

Steps for Completing the Square

Subtract 4x^{2} from both sides.

$$8x+40-4x^{2}=-56$$

Subtract 40 from both sides.

$$8x-4x^{2}=-56-40$$

Subtract 40 from -56 to get -96.

$$8x-4x^{2}=-96$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$-4x^{2}+8x=-96$$

Divide both sides by -4.

$$\frac{-4x^{2}+8x}{-4}=-\frac{96}{-4}$$

Dividing by -4 undoes the multiplication by -4.

$$x^{2}+\frac{8}{-4}x=-\frac{96}{-4}$$

Divide 8 by -4.

$$x^{2}-2x=-\frac{96}{-4}$$

Divide -96 by -4.

$$x^{2}-2x=24$$

Divide -2, the coefficient of the x term, by 2 to get -1. Then add the square of -1 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-2x+1=24+1$$

Add 24 to 1.

$$x^{2}-2x+1=25$$

Factor x^{2}-2x+1. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-1\right)^{2}=25$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-1\right)^{2}}=\sqrt{25}$$

Simplify.

$$x-1=5 x-1=-5$$

Add 1 to both sides of the equation.

$$x=6 x=-4$$