Solve 18x^2+33x=180

$$18 { x }^{ 2 } +33x=180$$

Solve for x

$$x = \frac{\sqrt{1561} - 11}{12} \approx 2.375791044$$

$$x=\frac{-\sqrt{1561}-11}{12}\approx -4.209124378$$

Steps Using the Quadratic Formula

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$18x^{2}+33x=180$$

Subtract 180 from both sides of the equation.

$$18x^{2}+33x-180=180-180$$

Subtracting 180 from itself leaves 0.

$$18x^{2}+33x-180=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 18 for a, 33 for b, and -180 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-33±\sqrt{33^{2}-4\times 18\left(-180\right)}}{2\times 18}$$

Square 33.

$$x=\frac{-33±\sqrt{1089-4\times 18\left(-180\right)}}{2\times 18}$$

Multiply -4 times 18.

$$x=\frac{-33±\sqrt{1089-72\left(-180\right)}}{2\times 18}$$

Multiply -72 times -180.

$$x=\frac{-33±\sqrt{1089+12960}}{2\times 18}$$

Add 1089 to 12960.

$$x=\frac{-33±\sqrt{14049}}{2\times 18}$$

Take the square root of 14049.

$$x=\frac{-33±3\sqrt{1561}}{2\times 18}$$

Multiply 2 times 18.

$$x=\frac{-33±3\sqrt{1561}}{36}$$

Now solve the equation x=\frac{-33±3\sqrt{1561}}{36} when ± is plus. Add -33 to 3\sqrt{1561}.

$$x=\frac{3\sqrt{1561}-33}{36}$$

Divide -33+3\sqrt{1561} by 36.

$$x=\frac{\sqrt{1561}-11}{12}$$

Now solve the equation x=\frac{-33±3\sqrt{1561}}{36} when ± is minus. Subtract 3\sqrt{1561} from -33.

$$x=\frac{-3\sqrt{1561}-33}{36}$$

Divide -33-3\sqrt{1561} by 36.

$$x=\frac{-\sqrt{1561}-11}{12}$$

The equation is now solved.

$$x=\frac{\sqrt{1561}-11}{12} x=\frac{-\sqrt{1561}-11}{12}$$

Steps for Completing the Square

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$18x^{2}+33x=180$$

Divide both sides by 18.

$$\frac{18x^{2}+33x}{18}=\frac{180}{18}$$

Dividing by 18 undoes the multiplication by 18.

$$x^{2}+\frac{33}{18}x=\frac{180}{18}$$

Reduce the fraction \frac{33}{18} to lowest terms by extracting and canceling out 3.

$$x^{2}+\frac{11}{6}x=\frac{180}{18}$$

Divide 180 by 18.

$$x^{2}+\frac{11}{6}x=10$$

Divide \frac{11}{6}, the coefficient of the x term, by 2 to get \frac{11}{12}. Then add the square of \frac{11}{12} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{11}{6}x+\left(\frac{11}{12}\right)^{2}=10+\left(\frac{11}{12}\right)^{2}$$

Square \frac{11}{12} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{11}{6}x+\frac{121}{144}=10+\frac{121}{144}$$

Add 10 to \frac{121}{144}.

$$x^{2}+\frac{11}{6}x+\frac{121}{144}=\frac{1561}{144}$$

Factor x^{2}+\frac{11}{6}x+\frac{121}{144}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{11}{12}\right)^{2}=\frac{1561}{144}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{11}{12}\right)^{2}}=\sqrt{\frac{1561}{144}}$$

Simplify.

$$x+\frac{11}{12}=\frac{\sqrt{1561}}{12} x+\frac{11}{12}=-\frac{\sqrt{1561}}{12}$$

Subtract \frac{11}{12} from both sides of the equation.

$$x=\frac{\sqrt{1561}-11}{12} x=\frac{-\sqrt{1561}-11}{12}$$