Solve 5x^2+15x=12x-13

$$5 { x }^{ 2 } +15x=12x-13$$

Solve for x (complex solution)

$$x=\frac{-3+\sqrt{251}i}{10}\approx -0.3+1.584297952i$$

$$x=\frac{-\sqrt{251}i-3}{10}\approx -0.3-1.584297952i$$

Steps Using the Quadratic Formula

Subtract 12x from both sides.

$$5x^{2}+15x-12x=-13$$

Combine 15x and -12x to get 3x.

$$5x^{2}+3x=-13$$

Add 13 to both sides.

$$5x^{2}+3x+13=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, 3 for b, and 13 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-3±\sqrt{3^{2}-4\times 5\times 13}}{2\times 5}$$

Square 3.

$$x=\frac{-3±\sqrt{9-4\times 5\times 13}}{2\times 5}$$

Multiply -4 times 5.

$$x=\frac{-3±\sqrt{9-20\times 13}}{2\times 5}$$

Multiply -20 times 13.

$$x=\frac{-3±\sqrt{9-260}}{2\times 5}$$

Add 9 to -260.

$$x=\frac{-3±\sqrt{-251}}{2\times 5}$$

Take the square root of -251.

$$x=\frac{-3±\sqrt{251}i}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{-3±\sqrt{251}i}{10}$$

Now solve the equation x=\frac{-3±\sqrt{251}i}{10} when ± is plus. Add -3 to i\sqrt{251}.

$$x=\frac{-3+\sqrt{251}i}{10}$$

Now solve the equation x=\frac{-3±\sqrt{251}i}{10} when ± is minus. Subtract i\sqrt{251} from -3.

$$x=\frac{-\sqrt{251}i-3}{10}$$

The equation is now solved.

$$x=\frac{-3+\sqrt{251}i}{10} x=\frac{-\sqrt{251}i-3}{10}$$

Steps for Completing the Square

Subtract 12x from both sides.

$$5x^{2}+15x-12x=-13$$

Combine 15x and -12x to get 3x.

$$5x^{2}+3x=-13$$

Divide both sides by 5.

$$\frac{5x^{2}+3x}{5}=-\frac{13}{5}$$

Dividing by 5 undoes the multiplication by 5.

$$x^{2}+\frac{3}{5}x=-\frac{13}{5}$$

Divide \frac{3}{5}, the coefficient of the x term, by 2 to get \frac{3}{10}. Then add the square of \frac{3}{10} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{3}{5}x+\left(\frac{3}{10}\right)^{2}=-\frac{13}{5}+\left(\frac{3}{10}\right)^{2}$$

Square \frac{3}{10} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{3}{5}x+\frac{9}{100}=-\frac{13}{5}+\frac{9}{100}$$

Add -\frac{13}{5} to \frac{9}{100} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{3}{5}x+\frac{9}{100}=-\frac{251}{100}$$

Factor x^{2}+\frac{3}{5}x+\frac{9}{100}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{3}{10}\right)^{2}=-\frac{251}{100}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{3}{10}\right)^{2}}=\sqrt{-\frac{251}{100}}$$

Simplify.

$$x+\frac{3}{10}=\frac{\sqrt{251}i}{10} x+\frac{3}{10}=-\frac{\sqrt{251}i}{10}$$

Subtract \frac{3}{10} from both sides of the equation.

$$x=\frac{-3+\sqrt{251}i}{10} x=\frac{-\sqrt{251}i-3}{10}$$