Solve 6x^2+15x=11x-3

$$6 x ^ { 2 } + 15 x = 11 x - 3$$

Solve for x (complex solution)

$$x=\frac{\sqrt{14}i}{6}-\frac{1}{3}\approx -0.333333333+0.623609564i$$

$$x=-\frac{\sqrt{14}i}{6}-\frac{1}{3}\approx -0.333333333-0.623609564i$$

Steps Using the Quadratic Formula

Subtract 11x from both sides.

$$6x^{2}+15x-11x=-3$$

Combine 15x and -11x to get 4x.

$$6x^{2}+4x=-3$$

Add 3 to both sides.

$$6x^{2}+4x+3=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 6 for a, 4 for b, and 3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-4±\sqrt{4^{2}-4\times 6\times 3}}{2\times 6}$$

Square 4.

$$x=\frac{-4±\sqrt{16-4\times 6\times 3}}{2\times 6}$$

Multiply -4 times 6.

$$x=\frac{-4±\sqrt{16-24\times 3}}{2\times 6}$$

Multiply -24 times 3.

$$x=\frac{-4±\sqrt{16-72}}{2\times 6}$$

Add 16 to -72.

$$x=\frac{-4±\sqrt{-56}}{2\times 6}$$

Take the square root of -56.

$$x=\frac{-4±2\sqrt{14}i}{2\times 6}$$

Multiply 2 times 6.

$$x=\frac{-4±2\sqrt{14}i}{12}$$

Now solve the equation x=\frac{-4±2\sqrt{14}i}{12} when ± is plus. Add -4 to 2i\sqrt{14}.

$$x=\frac{-4+2\sqrt{14}i}{12}$$

Divide -4+2i\sqrt{14} by 12.

$$x=\frac{\sqrt{14}i}{6}-\frac{1}{3}$$

Now solve the equation x=\frac{-4±2\sqrt{14}i}{12} when ± is minus. Subtract 2i\sqrt{14} from -4.

$$x=\frac{-2\sqrt{14}i-4}{12}$$

Divide -4-2i\sqrt{14} by 12.

$$x=-\frac{\sqrt{14}i}{6}-\frac{1}{3}$$

The equation is now solved.

$$x=\frac{\sqrt{14}i}{6}-\frac{1}{3} x=-\frac{\sqrt{14}i}{6}-\frac{1}{3}$$

Steps for Completing the Square

Subtract 11x from both sides.

$$6x^{2}+15x-11x=-3$$

Combine 15x and -11x to get 4x.

$$6x^{2}+4x=-3$$

Divide both sides by 6.

$$\frac{6x^{2}+4x}{6}=-\frac{3}{6}$$

Dividing by 6 undoes the multiplication by 6.

$$x^{2}+\frac{4}{6}x=-\frac{3}{6}$$

Reduce the fraction \frac{4}{6} to lowest terms by extracting and canceling out 2.

$$x^{2}+\frac{2}{3}x=-\frac{3}{6}$$

Reduce the fraction \frac{-3}{6} to lowest terms by extracting and canceling out 3.

$$x^{2}+\frac{2}{3}x=-\frac{1}{2}$$

Divide \frac{2}{3}, the coefficient of the x term, by 2 to get \frac{1}{3}. Then add the square of \frac{1}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{2}{3}x+\left(\frac{1}{3}\right)^{2}=-\frac{1}{2}+\left(\frac{1}{3}\right)^{2}$$

Square \frac{1}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{2}{3}x+\frac{1}{9}=-\frac{1}{2}+\frac{1}{9}$$

Add -\frac{1}{2} to \frac{1}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{2}{3}x+\frac{1}{9}=-\frac{7}{18}$$

Factor x^{2}+\frac{2}{3}x+\frac{1}{9}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{1}{3}\right)^{2}=-\frac{7}{18}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{3}\right)^{2}}=\sqrt{-\frac{7}{18}}$$

Simplify.

$$x+\frac{1}{3}=\frac{\sqrt{14}i}{6} x+\frac{1}{3}=-\frac{\sqrt{14}i}{6}$$

Subtract \frac{1}{3} from both sides of the equation.

$$x=\frac{\sqrt{14}i}{6}-\frac{1}{3} x=-\frac{\sqrt{14}i}{6}-\frac{1}{3}$$