Solve -3x^2=-11x+12

$$- 3 x ^ { 2 } = - 11 x + 12$$

Solve for x (complex solution)

$$x=\frac{-\sqrt{23}i+11}{6}\approx 1.833333333-0.799305254i$$

$$x=\frac{11+\sqrt{23}i}{6}\approx 1.833333333+0.799305254i$$

Steps Using the Quadratic Formula

Add 11x to both sides.

$$-3x^{2}+11x=12$$

Subtract 12 from both sides.

$$-3x^{2}+11x-12=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -3 for a, 11 for b, and -12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-11±\sqrt{11^{2}-4\left(-3\right)\left(-12\right)}}{2\left(-3\right)}$$

Square 11.

$$x=\frac{-11±\sqrt{121-4\left(-3\right)\left(-12\right)}}{2\left(-3\right)}$$

Multiply -4 times -3.

$$x=\frac{-11±\sqrt{121+12\left(-12\right)}}{2\left(-3\right)}$$

Multiply 12 times -12.

$$x=\frac{-11±\sqrt{121-144}}{2\left(-3\right)}$$

Add 121 to -144.

$$x=\frac{-11±\sqrt{-23}}{2\left(-3\right)}$$

Take the square root of -23.

$$x=\frac{-11±\sqrt{23}i}{2\left(-3\right)}$$

Multiply 2 times -3.

$$x=\frac{-11±\sqrt{23}i}{-6}$$

Now solve the equation x=\frac{-11±\sqrt{23}i}{-6} when ± is plus. Add -11 to i\sqrt{23}.

$$x=\frac{-11+\sqrt{23}i}{-6}$$

Divide -11+i\sqrt{23} by -6.

$$x=\frac{-\sqrt{23}i+11}{6}$$

Now solve the equation x=\frac{-11±\sqrt{23}i}{-6} when ± is minus. Subtract i\sqrt{23} from -11.

$$x=\frac{-\sqrt{23}i-11}{-6}$$

Divide -11-i\sqrt{23} by -6.

$$x=\frac{11+\sqrt{23}i}{6}$$

The equation is now solved.

$$x=\frac{-\sqrt{23}i+11}{6} x=\frac{11+\sqrt{23}i}{6}$$

Steps for Completing the Square

Add 11x to both sides.

$$-3x^{2}+11x=12$$

Divide both sides by -3.

$$\frac{-3x^{2}+11x}{-3}=\frac{12}{-3}$$

Dividing by -3 undoes the multiplication by -3.

$$x^{2}+\frac{11}{-3}x=\frac{12}{-3}$$

Divide 11 by -3.

$$x^{2}-\frac{11}{3}x=\frac{12}{-3}$$

Divide 12 by -3.

$$x^{2}-\frac{11}{3}x=-4$$

Divide -\frac{11}{3}, the coefficient of the x term, by 2 to get -\frac{11}{6}. Then add the square of -\frac{11}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{11}{3}x+\left(-\frac{11}{6}\right)^{2}=-4+\left(-\frac{11}{6}\right)^{2}$$

Square -\frac{11}{6} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{11}{3}x+\frac{121}{36}=-4+\frac{121}{36}$$

Add -4 to \frac{121}{36}.

$$x^{2}-\frac{11}{3}x+\frac{121}{36}=-\frac{23}{36}$$

Factor x^{2}-\frac{11}{3}x+\frac{121}{36}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{11}{6}\right)^{2}=-\frac{23}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{11}{6}\right)^{2}}=\sqrt{-\frac{23}{36}}$$

Simplify.

$$x-\frac{11}{6}=\frac{\sqrt{23}i}{6} x-\frac{11}{6}=-\frac{\sqrt{23}i}{6}$$

Add \frac{11}{6} to both sides of the equation.

$$x=\frac{11+\sqrt{23}i}{6} x=\frac{-\sqrt{23}i+11}{6}$$