Solve (x+3)x=40

$$( x + 3 ) x = 40$$

Solve for x

$$x=-8$$

$$x=5$$

Steps Using the Quadratic Formula

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

$$x^{2}+3x=40$$

Subtract 40 from both sides.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 3 for b, and -40 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-3±\sqrt{3^{2}-4\left(-40\right)}}{2}$$

Square 3.

$$x=\frac{-3±\sqrt{9-4\left(-40\right)}}{2}$$

Multiply -4 times -40.

$$x=\frac{-3±\sqrt{9+160}}{2}$$

Add 9 to 160.

$$x=\frac{-3±\sqrt{169}}{2}$$

Take the square root of 169.

$$x=\frac{-3±13}{2}$$

Now solve the equation x=\frac{-3±13}{2} when ± is plus. Add -3 to 13.

$$x=\frac{10}{2}$$

Divide 10 by 2.

$$x=5$$

Now solve the equation x=\frac{-3±13}{2} when ± is minus. Subtract 13 from -3.

$$x=-\frac{16}{2}$$

Divide -16 by 2.

$$x=-8$$

The equation is now solved.

$$x=5 x=-8$$

Steps for Completing the Square

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

$$x^{2}+3x=40$$

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

$$x^{2}+3x+\left(\frac{3}{2}\right)^{2}=40+\left(\frac{3}{2}\right)^{2}$$

Square \frac{3}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+3x+\frac{9}{4}=40+\frac{9}{4}$$

Add 40 to \frac{9}{4}.

$$x^{2}+3x+\frac{9}{4}=\frac{169}{4}$$

Factor x^{2}+3x+\frac{9}{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}.

$$\left(x+\frac{3}{2}\right)^{2}=\frac{169}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{3}{2}\right)^{2}}=\sqrt{\frac{169}{4}}$$

Simplify.

$$x+\frac{3}{2}=\frac{13}{2} x+\frac{3}{2}=-\frac{13}{2}$$

Subtract \frac{3}{2} from both sides of the equation.

$$x=5 x=-8$$