Solve x*left(x+3right)=18

$$x \times \left( x+3 \right) = 18$$

Solve for x

$$x=-6$$

$$x=3$$

Steps Using the Quadratic Formula

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

$$x^{2}+3x=18$$

Subtract 18 from both sides.

$$x^{2}+3x-18=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 3 for b, and -18 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-3±\sqrt{3^{2}-4\left(-18\right)}}{2}$$

Square 3.

$$x=\frac{-3±\sqrt{9-4\left(-18\right)}}{2}$$

Multiply -4 times -18.

$$x=\frac{-3±\sqrt{9+72}}{2}$$

Add 9 to 72.

$$x=\frac{-3±\sqrt{81}}{2}$$

Take the square root of 81.

$$x=\frac{-3±9}{2}$$

Now solve the equation x=\frac{-3±9}{2} when ± is plus. Add -3 to 9.

$$x=\frac{6}{2}$$

Divide 6 by 2.

$$x=3$$

Now solve the equation x=\frac{-3±9}{2} when ± is minus. Subtract 9 from -3.

$$x=-\frac{12}{2}$$

Divide -12 by 2.

$$x=-6$$

The equation is now solved.

$$x=3 x=-6$$

Steps for Completing the Square

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

$$x^{2}+3x=18$$

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.

$$x^{2}+3x+\left(\frac{3}{2}\right)^{2}=18+\left(\frac{3}{2}\right)^{2}$$

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

$$x^{2}+3x+\frac{9}{4}=18+\frac{9}{4}$$

Add 18 to \frac{9}{4}.

$$x^{2}+3x+\frac{9}{4}=\frac{81}{4}$$

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}.

$$\left(x+\frac{3}{2}\right)^{2}=\frac{81}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{3}{2}\right)^{2}}=\sqrt{\frac{81}{4}}$$

Simplify.

$$x+\frac{3}{2}=\frac{9}{2} x+\frac{3}{2}=-\frac{9}{2}$$

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

$$x=3 x=-6$$