Solve x^2=15x+16

$$x ^ { 2 } = 15 x + 16$$

Solve for x

$$x=-1$$

$$x=16$$

Steps Using Factoring

Subtract 15x from both sides.

$$x^{2}-15x=16$$

Subtract 16 from both sides.

$$x^{2}-15x-16=0$$

To solve the equation, factor x^{2}-15x-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.

$$a+b=-15 ab=-16$$

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 -16.

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

Calculate the sum for each pair.

$$1-16=-15 2-8=-6 4-4=0$$

The solution is the pair that gives sum -15.

$$a=-16 b=1$$

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

$$\left(x-16\right)\left(x+1\right)$$

To find equation solutions, solve x-16=0 and x+1=0.

$$x=16 x=-1$$

Steps Using Factoring By Grouping

Subtract 15x from both sides.

$$x^{2}-15x=16$$

Subtract 16 from both sides.

$$x^{2}-15x-16=0$$

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.

$$a+b=-15 ab=1\left(-16\right)=-16$$

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 -16.

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

Calculate the sum for each pair.

$$1-16=-15 2-8=-6 4-4=0$$

The solution is the pair that gives sum -15.

$$a=-16 b=1$$

Rewrite x^{2}-15x-16 as \left(x^{2}-16x\right)+\left(x-16\right).

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

Factor out x in x^{2}-16x.

$$x\left(x-16\right)+x-16$$

Factor out common term x-16 by using distributive property.

$$\left(x-16\right)\left(x+1\right)$$

To find equation solutions, solve x-16=0 and x+1=0.

$$x=16 x=-1$$

Steps Using the Quadratic Formula

Subtract 15x from both sides.

$$x^{2}-15x=16$$

Subtract 16 from both sides.

$$x^{2}-15x-16=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -15 for b, and -16 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-15\right)±\sqrt{\left(-15\right)^{2}-4\left(-16\right)}}{2}$$

Square -15.

$$x=\frac{-\left(-15\right)±\sqrt{225-4\left(-16\right)}}{2}$$

Multiply -4 times -16.

$$x=\frac{-\left(-15\right)±\sqrt{225+64}}{2}$$

Add 225 to 64.

$$x=\frac{-\left(-15\right)±\sqrt{289}}{2}$$

Take the square root of 289.

$$x=\frac{-\left(-15\right)±17}{2}$$

The opposite of -15 is 15.

$$x=\frac{15±17}{2}$$

Now solve the equation x=\frac{15±17}{2} when ± is plus. Add 15 to 17.

$$x=\frac{32}{2}$$

Divide 32 by 2.

$$x=16$$

Now solve the equation x=\frac{15±17}{2} when ± is minus. Subtract 17 from 15.

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

Divide -2 by 2.

$$x=-1$$

The equation is now solved.

$$x=16 x=-1$$

Steps for Completing the Square

Subtract 15x from both sides.

$$x^{2}-15x=16$$

Divide -15, the coefficient of the x term, by 2 to get -\frac{15}{2}. Then add the square of -\frac{15}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-15x+\left(-\frac{15}{2}\right)^{2}=16+\left(-\frac{15}{2}\right)^{2}$$

Square -\frac{15}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-15x+\frac{225}{4}=16+\frac{225}{4}$$

Add 16 to \frac{225}{4}.

$$x^{2}-15x+\frac{225}{4}=\frac{289}{4}$$

Factor x^{2}-15x+\frac{225}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{15}{2}\right)^{2}=\frac{289}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{15}{2}\right)^{2}}=\sqrt{\frac{289}{4}}$$

Simplify.

$$x-\frac{15}{2}=\frac{17}{2} x-\frac{15}{2}=-\frac{17}{2}$$

Add \frac{15}{2} to both sides of the equation.

$$x=16 x=-1$$