Solve 9=8x-6x^2

$$9 = 8 x - 6 x ^ { 2 }$$

Solve for x (complex solution)

$$x=\frac{\sqrt{38}i}{6}+\frac{2}{3}\approx 0.666666667+1.027402334i$$

$$x=-\frac{\sqrt{38}i}{6}+\frac{2}{3}\approx 0.666666667-1.027402334i$$

Steps Using the Quadratic Formula

Swap sides so that all variable terms are on the left hand side.

$$8x-6x^{2}=9$$

Subtract 9 from both sides.

$$8x-6x^{2}-9=0$$

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.

$$-6x^{2}+8x-9=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -6 for a, 8 for b, and -9 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-8±\sqrt{8^{2}-4\left(-6\right)\left(-9\right)}}{2\left(-6\right)}$$

Square 8.

$$x=\frac{-8±\sqrt{64-4\left(-6\right)\left(-9\right)}}{2\left(-6\right)}$$

Multiply -4 times -6.

$$x=\frac{-8±\sqrt{64+24\left(-9\right)}}{2\left(-6\right)}$$

Multiply 24 times -9.

$$x=\frac{-8±\sqrt{64-216}}{2\left(-6\right)}$$

Add 64 to -216.

$$x=\frac{-8±\sqrt{-152}}{2\left(-6\right)}$$

Take the square root of -152.

$$x=\frac{-8±2\sqrt{38}i}{2\left(-6\right)}$$

Multiply 2 times -6.

$$x=\frac{-8±2\sqrt{38}i}{-12}$$

Now solve the equation x=\frac{-8±2\sqrt{38}i}{-12} when ± is plus. Add -8 to 2i\sqrt{38}.

$$x=\frac{-8+2\sqrt{38}i}{-12}$$

Divide -8+2i\sqrt{38} by -12.

$$x=-\frac{\sqrt{38}i}{6}+\frac{2}{3}$$

Now solve the equation x=\frac{-8±2\sqrt{38}i}{-12} when ± is minus. Subtract 2i\sqrt{38} from -8.

$$x=\frac{-2\sqrt{38}i-8}{-12}$$

Divide -8-2i\sqrt{38} by -12.

$$x=\frac{\sqrt{38}i}{6}+\frac{2}{3}$$

The equation is now solved.

$$x=-\frac{\sqrt{38}i}{6}+\frac{2}{3} x=\frac{\sqrt{38}i}{6}+\frac{2}{3}$$

Steps for Completing the Square

Swap sides so that all variable terms are on the left hand side.

$$8x-6x^{2}=9$$

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.

$$-6x^{2}+8x=9$$

Divide both sides by -6.

$$\frac{-6x^{2}+8x}{-6}=\frac{9}{-6}$$

Dividing by -6 undoes the multiplication by -6.

$$x^{2}+\frac{8}{-6}x=\frac{9}{-6}$$

Reduce the fraction \frac{8}{-6} to lowest terms by extracting and canceling out 2.

$$x^{2}-\frac{4}{3}x=\frac{9}{-6}$$

Reduce the fraction \frac{9}{-6} to lowest terms by extracting and canceling out 3.

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

Divide -\frac{4}{3}, the coefficient of the x term, by 2 to get -\frac{2}{3}. Then add the square of -\frac{2}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

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

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

Add -\frac{3}{2} to \frac{4}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{4}{3}x+\frac{4}{9}=-\frac{19}{18}$$

Factor x^{2}-\frac{4}{3}x+\frac{4}{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}.

$$\left(x-\frac{2}{3}\right)^{2}=-\frac{19}{18}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{2}{3}\right)^{2}}=\sqrt{-\frac{19}{18}}$$

Simplify.

$$x-\frac{2}{3}=\frac{\sqrt{38}i}{6} x-\frac{2}{3}=-\frac{\sqrt{38}i}{6}$$

Add \frac{2}{3} to both sides of the equation.

$$x=\frac{\sqrt{38}i}{6}+\frac{2}{3} x=-\frac{\sqrt{38}i}{6}+\frac{2}{3}$$