Solve (x+5)^2-3(x+5)-10=0

$$( x + 5 ) ^ { 2 } - 3 ( x + 5 ) - 10 = 0$$

Solve for x

$$x=-7$$

$$x=0$$

Steps Using Factoring

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+5\right)^{2}.

$$x^{2}+10x+25-3\left(x+5\right)-10=0$$

Use the distributive property to multiply -3 by x+5.

$$x^{2}+10x+25-3x-15-10=0$$

Combine 10x and -3x to get 7x.

$$x^{2}+7x+25-15-10=0$$

Subtract 15 from 25 to get 10.

$$x^{2}+7x+10-10=0$$

Subtract 10 from 10 to get 0.

$$x^{2}+7x=0$$

Factor out x.

$$x\left(x+7\right)=0$$

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

$$x=0 x=-7$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+5\right)^{2}.

$$x^{2}+10x+25-3\left(x+5\right)-10=0$$

Use the distributive property to multiply -3 by x+5.

$$x^{2}+10x+25-3x-15-10=0$$

Combine 10x and -3x to get 7x.

$$x^{2}+7x+25-15-10=0$$

Subtract 15 from 25 to get 10.

$$x^{2}+7x+10-10=0$$

Subtract 10 from 10 to get 0.

$$x^{2}+7x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 7 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-7±\sqrt{7^{2}}}{2}$$

Take the square root of 7^{2}.

$$x=\frac{-7±7}{2}$$

Now solve the equation x=\frac{-7±7}{2} when ± is plus. Add -7 to 7.

$$x=\frac{0}{2}$$

Divide 0 by 2.

$$x=0$$

Now solve the equation x=\frac{-7±7}{2} when ± is minus. Subtract 7 from -7.

$$x=-\frac{14}{2}$$

Divide -14 by 2.

$$x=-7$$

The equation is now solved.

$$x=0 x=-7$$

Steps for Completing the Square

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+5\right)^{2}.

$$x^{2}+10x+25-3\left(x+5\right)-10=0$$

Use the distributive property to multiply -3 by x+5.

$$x^{2}+10x+25-3x-15-10=0$$

Combine 10x and -3x to get 7x.

$$x^{2}+7x+25-15-10=0$$

Subtract 15 from 25 to get 10.

$$x^{2}+7x+10-10=0$$

Subtract 10 from 10 to get 0.

$$x^{2}+7x=0$$

Divide 7, the coefficient of the x term, by 2 to get \frac{7}{2}. Then add the square of \frac{7}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+7x+\left(\frac{7}{2}\right)^{2}=\left(\frac{7}{2}\right)^{2}$$

Square \frac{7}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+7x+\frac{49}{4}=\frac{49}{4}$$

Factor x^{2}+7x+\frac{49}{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}.

$$\left(x+\frac{7}{2}\right)^{2}=\frac{49}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{7}{2}\right)^{2}}=\sqrt{\frac{49}{4}}$$

Simplify.

$$x+\frac{7}{2}=\frac{7}{2} x+\frac{7}{2}=-\frac{7}{2}$$

Subtract \frac{7}{2} from both sides of the equation.

$$x=0 x=-7$$