Solve 26=586-32x-16x^2

$$26=586-32x-16 { x }^{ 2 }$$

Solve for x

$$x=-7$$

$$x=5$$

Steps Using Factoring By Grouping

Swap sides so that all variable terms are on the left hand side.

$$586-32x-16x^{2}=26$$

Subtract 26 from both sides.

$$586-32x-16x^{2}-26=0$$

Subtract 26 from 586 to get 560.

$$560-32x-16x^{2}=0$$

Divide both sides by 16.

$$35-2x-x^{2}=0$$

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

$$-x^{2}-2x+35=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx+35. To find a and b, set up a system to be solved.

$$a+b=-2 ab=-35=-35$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -35.

$$1,-35 5,-7$$

Calculate the sum for each pair.

$$1-35=-34 5-7=-2$$

The solution is the pair that gives sum -2.

$$a=5 b=-7$$

Rewrite -x^{2}-2x+35 as \left(-x^{2}+5x\right)+\left(-7x+35\right).

$$\left(-x^{2}+5x\right)+\left(-7x+35\right)$$

Factor out x in the first and 7 in the second group.

$$x\left(-x+5\right)+7\left(-x+5\right)$$

Factor out common term -x+5 by using distributive property.

$$\left(-x+5\right)\left(x+7\right)$$

To find equation solutions, solve -x+5=0 and x+7=0.

$$x=5 x=-7$$

Steps Using the Quadratic Formula

Swap sides so that all variable terms are on the left hand side.

$$586-32x-16x^{2}=26$$

Subtract 26 from both sides.

$$586-32x-16x^{2}-26=0$$

Subtract 26 from 586 to get 560.

$$560-32x-16x^{2}=0$$

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.

$$-16x^{2}-32x+560=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -16 for a, -32 for b, and 560 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-32\right)±\sqrt{\left(-32\right)^{2}-4\left(-16\right)\times 560}}{2\left(-16\right)}$$

Square -32.

$$x=\frac{-\left(-32\right)±\sqrt{1024-4\left(-16\right)\times 560}}{2\left(-16\right)}$$

Multiply -4 times -16.

$$x=\frac{-\left(-32\right)±\sqrt{1024+64\times 560}}{2\left(-16\right)}$$

Multiply 64 times 560.

$$x=\frac{-\left(-32\right)±\sqrt{1024+35840}}{2\left(-16\right)}$$

Add 1024 to 35840.

$$x=\frac{-\left(-32\right)±\sqrt{36864}}{2\left(-16\right)}$$

Take the square root of 36864.

$$x=\frac{-\left(-32\right)±192}{2\left(-16\right)}$$

The opposite of -32 is 32.

$$x=\frac{32±192}{2\left(-16\right)}$$

Multiply 2 times -16.

$$x=\frac{32±192}{-32}$$

Now solve the equation x=\frac{32±192}{-32} when ± is plus. Add 32 to 192.

$$x=\frac{224}{-32}$$

Divide 224 by -32.

$$x=-7$$

Now solve the equation x=\frac{32±192}{-32} when ± is minus. Subtract 192 from 32.

$$x=-\frac{160}{-32}$$

Divide -160 by -32.

$$x=5$$

The equation is now solved.

$$x=-7 x=5$$

Steps for Completing the Square

Swap sides so that all variable terms are on the left hand side.

$$586-32x-16x^{2}=26$$

Subtract 586 from both sides.

$$-32x-16x^{2}=26-586$$

Subtract 586 from 26 to get -560.

$$-32x-16x^{2}=-560$$

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.

$$-16x^{2}-32x=-560$$

Divide both sides by -16.

$$\frac{-16x^{2}-32x}{-16}=-\frac{560}{-16}$$

Dividing by -16 undoes the multiplication by -16.

$$x^{2}+\left(-\frac{32}{-16}\right)x=-\frac{560}{-16}$$

Divide -32 by -16.

$$x^{2}+2x=-\frac{560}{-16}$$

Divide -560 by -16.

$$x^{2}+2x=35$$

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.

$$x^{2}+2x+1^{2}=35+1^{2}$$

Square 1.

$$x^{2}+2x+1=35+1$$

Add 35 to 1.

$$x^{2}+2x+1=36$$

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}.

$$\left(x+1\right)^{2}=36$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+1\right)^{2}}=\sqrt{36}$$

Simplify.

$$x+1=6 x+1=-6$$

Subtract 1 from both sides of the equation.

$$x=5 x=-7$$