Solve 3x=2x^2-3

$$3 x = 2 x ^ { 2 } - 3$$

Solve for x

$$x = \frac{\sqrt{33} + 3}{4} \approx 2.186140662$$

$$x=\frac{3-\sqrt{33}}{4}\approx -0.686140662$$

Steps Using the Quadratic Formula

Subtract 2x^{2} from both sides.

$$3x-2x^{2}=-3$$

Add 3 to both sides.

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

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.

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

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

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

Square 3.

$$x=\frac{-3±\sqrt{9-4\left(-2\right)\times 3}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$x=\frac{-3±\sqrt{9+8\times 3}}{2\left(-2\right)}$$

Multiply 8 times 3.

$$x=\frac{-3±\sqrt{9+24}}{2\left(-2\right)}$$

Add 9 to 24.

$$x=\frac{-3±\sqrt{33}}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{-3±\sqrt{33}}{-4}$$

Now solve the equation x=\frac{-3±\sqrt{33}}{-4} when ± is plus. Add -3 to \sqrt{33}.

$$x=\frac{\sqrt{33}-3}{-4}$$

Divide -3+\sqrt{33} by -4.

$$x=\frac{3-\sqrt{33}}{4}$$

Now solve the equation x=\frac{-3±\sqrt{33}}{-4} when ± is minus. Subtract \sqrt{33} from -3.

$$x=\frac{-\sqrt{33}-3}{-4}$$

Divide -3-\sqrt{33} by -4.

$$x=\frac{\sqrt{33}+3}{4}$$

The equation is now solved.

$$x=\frac{3-\sqrt{33}}{4} x=\frac{\sqrt{33}+3}{4}$$

Steps for Completing the Square

Subtract 2x^{2} from both sides.

$$3x-2x^{2}=-3$$

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.

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

Divide both sides by -2.

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

Dividing by -2 undoes the multiplication by -2.

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

Divide 3 by -2.

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

Divide -3 by -2.

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

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

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

Square -\frac{3}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{3}{2}x+\frac{9}{16}=\frac{3}{2}+\frac{9}{16}$$

Add \frac{3}{2} to \frac{9}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{3}{2}x+\frac{9}{16}=\frac{33}{16}$$

Factor x^{2}-\frac{3}{2}x+\frac{9}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{3}{4}\right)^{2}=\frac{33}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3}{4}\right)^{2}}=\sqrt{\frac{33}{16}}$$

Simplify.

$$x-\frac{3}{4}=\frac{\sqrt{33}}{4} x-\frac{3}{4}=-\frac{\sqrt{33}}{4}$$

Add \frac{3}{4} to both sides of the equation.

$$x=\frac{\sqrt{33}+3}{4} x=\frac{3-\sqrt{33}}{4}$$