Solve -3x=2x^2-4

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

Solve for x

$$x=\frac{\sqrt{41}-3}{4}\approx 0.850781059$$

$$x=\frac{-\sqrt{41}-3}{4}\approx -2.350781059$$

Steps Using the Quadratic Formula

Subtract 2x^{2} from both sides.

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

Add 4 to both sides.

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

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.

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

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

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

Square -3.

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

Multiply -4 times -2.

$$x=\frac{-\left(-3\right)±\sqrt{9+8\times 4}}{2\left(-2\right)}$$

Multiply 8 times 4.

$$x=\frac{-\left(-3\right)±\sqrt{9+32}}{2\left(-2\right)}$$

Add 9 to 32.

$$x=\frac{-\left(-3\right)±\sqrt{41}}{2\left(-2\right)}$$

The opposite of -3 is 3.

$$x=\frac{3±\sqrt{41}}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{3±\sqrt{41}}{-4}$$

Now solve the equation x=\frac{3±\sqrt{41}}{-4} when ± is plus. Add 3 to \sqrt{41}.

$$x=\frac{\sqrt{41}+3}{-4}$$

Divide 3+\sqrt{41} by -4.

$$x=\frac{-\sqrt{41}-3}{4}$$

Now solve the equation x=\frac{3±\sqrt{41}}{-4} when ± is minus. Subtract \sqrt{41} from 3.

$$x=\frac{3-\sqrt{41}}{-4}$$

Divide 3-\sqrt{41} by -4.

$$x=\frac{\sqrt{41}-3}{4}$$

The equation is now solved.

$$x=\frac{-\sqrt{41}-3}{4} x=\frac{\sqrt{41}-3}{4}$$

Steps for Completing the Square

Subtract 2x^{2} from both sides.

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

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.

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

Divide both sides by -2.

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

Dividing by -2 undoes the multiplication by -2.

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

Divide -3 by -2.

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

Divide -4 by -2.

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

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.

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

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

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

Add 2 to \frac{9}{16}.

$$x^{2}+\frac{3}{2}x+\frac{9}{16}=\frac{41}{16}$$

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}.

$$\left(x+\frac{3}{4}\right)^{2}=\frac{41}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{3}{4}\right)^{2}}=\sqrt{\frac{41}{16}}$$

Simplify.

$$x+\frac{3}{4}=\frac{\sqrt{41}}{4} x+\frac{3}{4}=-\frac{\sqrt{41}}{4}$$

Subtract \frac{3}{4} from both sides of the equation.

$$x=\frac{\sqrt{41}-3}{4} x=\frac{-\sqrt{41}-3}{4}$$