Solve 3(x+5)^2-7(x+5)-40=0

$$3 ( x + 5 ) ^ { 2 } - 7 ( x + 5 ) - 40 = 0$$

Solve for x

$$x = -\frac{23}{3} = -7\frac{2}{3} \approx -7.666666667$$

$$x=0$$

Steps Using Factoring

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

$$3\left(x^{2}+10x+25\right)-7\left(x+5\right)-40=0$$

Use the distributive property to multiply 3 by x^{2}+10x+25.

$$3x^{2}+30x+75-7\left(x+5\right)-40=0$$

Use the distributive property to multiply -7 by x+5.

$$3x^{2}+30x+75-7x-35-40=0$$

Combine 30x and -7x to get 23x.

$$3x^{2}+23x+75-35-40=0$$

Subtract 35 from 75 to get 40.

$$3x^{2}+23x+40-40=0$$

Subtract 40 from 40 to get 0.

$$3x^{2}+23x=0$$

Factor out x.

$$x\left(3x+23\right)=0$$

To find equation solutions, solve x=0 and 3x+23=0.

$$x=0 x=-\frac{23}{3}$$

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}.

$$3\left(x^{2}+10x+25\right)-7\left(x+5\right)-40=0$$

Use the distributive property to multiply 3 by x^{2}+10x+25.

$$3x^{2}+30x+75-7\left(x+5\right)-40=0$$

Use the distributive property to multiply -7 by x+5.

$$3x^{2}+30x+75-7x-35-40=0$$

Combine 30x and -7x to get 23x.

$$3x^{2}+23x+75-35-40=0$$

Subtract 35 from 75 to get 40.

$$3x^{2}+23x+40-40=0$$

Subtract 40 from 40 to get 0.

$$3x^{2}+23x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, 23 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-23±\sqrt{23^{2}}}{2\times 3}$$

Take the square root of 23^{2}.

$$x=\frac{-23±23}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{-23±23}{6}$$

Now solve the equation x=\frac{-23±23}{6} when ± is plus. Add -23 to 23.

$$x=\frac{0}{6}$$

Divide 0 by 6.

$$x=0$$

Now solve the equation x=\frac{-23±23}{6} when ± is minus. Subtract 23 from -23.

$$x=-\frac{46}{6}$$

Reduce the fraction \frac{-46}{6} to lowest terms by extracting and canceling out 2.

$$x=-\frac{23}{3}$$

The equation is now solved.

$$x=0 x=-\frac{23}{3}$$

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}.

$$3\left(x^{2}+10x+25\right)-7\left(x+5\right)-40=0$$

Use the distributive property to multiply 3 by x^{2}+10x+25.

$$3x^{2}+30x+75-7\left(x+5\right)-40=0$$

Use the distributive property to multiply -7 by x+5.

$$3x^{2}+30x+75-7x-35-40=0$$

Combine 30x and -7x to get 23x.

$$3x^{2}+23x+75-35-40=0$$

Subtract 35 from 75 to get 40.

$$3x^{2}+23x+40-40=0$$

Subtract 40 from 40 to get 0.

$$3x^{2}+23x=0$$

Divide both sides by 3.

$$\frac{3x^{2}+23x}{3}=\frac{0}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}+\frac{23}{3}x=\frac{0}{3}$$

Divide 0 by 3.

$$x^{2}+\frac{23}{3}x=0$$

Divide \frac{23}{3}, the coefficient of the x term, by 2 to get \frac{23}{6}. Then add the square of \frac{23}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{23}{3}x+\left(\frac{23}{6}\right)^{2}=\left(\frac{23}{6}\right)^{2}$$

Square \frac{23}{6} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{23}{3}x+\frac{529}{36}=\frac{529}{36}$$

Factor x^{2}+\frac{23}{3}x+\frac{529}{36}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{23}{6}\right)^{2}=\frac{529}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{23}{6}\right)^{2}}=\sqrt{\frac{529}{36}}$$

Simplify.

$$x+\frac{23}{6}=\frac{23}{6} x+\frac{23}{6}=-\frac{23}{6}$$

Subtract \frac{23}{6} from both sides of the equation.

$$x=0 x=-\frac{23}{3}$$