Solve -16=x^2-8x

$$- 16 = x ^ { 2 } - 8 x$$

Solve for x

$$x=4$$

Steps Using Factoring

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

$$x^{2}-8x=-16$$

Add 16 to both sides.

$$x^{2}-8x+16=0$$

To solve the equation, factor x^{2}-8x+16 using formula x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right). To find a and b, set up a system to be solved.

$$a+b=-8 ab=16$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 16.

$$-1,-16 -2,-8 -4,-4$$

Calculate the sum for each pair.

$$-1-16=-17 -2-8=-10 -4-4=-8$$

The solution is the pair that gives sum -8.

$$a=-4 b=-4$$

Rewrite factored expression \left(x+a\right)\left(x+b\right) using the obtained values.

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

Rewrite as a binomial square.

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

To find equation solution, solve x-4=0.

$$x=4$$

Steps Using Factoring By Grouping

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

$$x^{2}-8x=-16$$

Add 16 to both sides.

$$x^{2}-8x+16=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx+16. To find a and b, set up a system to be solved.

$$a+b=-8 ab=1\times 16=16$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 16.

$$-1,-16 -2,-8 -4,-4$$

Calculate the sum for each pair.

$$-1-16=-17 -2-8=-10 -4-4=-8$$

The solution is the pair that gives sum -8.

$$a=-4 b=-4$$

Rewrite x^{2}-8x+16 as \left(x^{2}-4x\right)+\left(-4x+16\right).

$$\left(x^{2}-4x\right)+\left(-4x+16\right)$$

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

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

Factor out common term x-4 by using distributive property.

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

Rewrite as a binomial square.

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

To find equation solution, solve x-4=0.

$$x=4$$

Steps Using the Quadratic Formula

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

$$x^{2}-8x=-16$$

Add 16 to both sides.

$$x^{2}-8x+16=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -8 for b, and 16 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Square -8.

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

Multiply -4 times 16.

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

Add 64 to -64.

$$x=\frac{-\left(-8\right)±\sqrt{0}}{2}$$

Take the square root of 0.

$$x=-\frac{-8}{2}$$

The opposite of -8 is 8.

$$x=\frac{8}{2}$$

Divide 8 by 2.

$$x=4$$

Steps for Completing the Square

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

$$x^{2}-8x=-16$$

Divide -8, the coefficient of the x term, by 2 to get -4. Then add the square of -4 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-8x+\left(-4\right)^{2}=-16+\left(-4\right)^{2}$$

Square -4.

$$x^{2}-8x+16=-16+16$$

Add -16 to 16.

$$x^{2}-8x+16=0$$

Factor x^{2}-8x+16. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-4\right)^{2}=0$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-4\right)^{2}}=\sqrt{0}$$

Simplify.

$$x-4=0 x-4=0$$

Add 4 to both sides of the equation.

$$x=4 x=4$$

The equation is now solved. Solutions are the same.

$$x=4$$