Solve 0=-2x^2-4x+3

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

Solve for x

$$x=\frac{\sqrt{10}}{2}-1\approx 0.58113883$$

$$x=-\frac{\sqrt{10}}{2}-1\approx -2.58113883$$

Steps Using the Quadratic Formula

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

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

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

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

Square -4.

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

Multiply -4 times -2.

$$x=\frac{-\left(-4\right)±\sqrt{16+8\times 3}}{2\left(-2\right)}$$

Multiply 8 times 3.

$$x=\frac{-\left(-4\right)±\sqrt{16+24}}{2\left(-2\right)}$$

Add 16 to 24.

$$x=\frac{-\left(-4\right)±\sqrt{40}}{2\left(-2\right)}$$

Take the square root of 40.

$$x=\frac{-\left(-4\right)±2\sqrt{10}}{2\left(-2\right)}$$

The opposite of -4 is 4.

$$x=\frac{4±2\sqrt{10}}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{4±2\sqrt{10}}{-4}$$

Now solve the equation x=\frac{4±2\sqrt{10}}{-4} when ± is plus. Add 4 to 2\sqrt{10}.

$$x=\frac{2\sqrt{10}+4}{-4}$$

Divide 4+2\sqrt{10} by -4.

$$x=-\frac{\sqrt{10}}{2}-1$$

Now solve the equation x=\frac{4±2\sqrt{10}}{-4} when ± is minus. Subtract 2\sqrt{10} from 4.

$$x=\frac{4-2\sqrt{10}}{-4}$$

Divide 4-2\sqrt{10} by -4.

$$x=\frac{\sqrt{10}}{2}-1$$

The equation is now solved.

$$x=-\frac{\sqrt{10}}{2}-1 x=\frac{\sqrt{10}}{2}-1$$

Steps for Completing the Square

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

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

Subtract 3 from both sides. Anything subtracted from zero gives its negation.

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

Divide both sides by -2.

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

Dividing by -2 undoes the multiplication by -2.

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

Divide -4 by -2.

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

Divide -3 by -2.

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

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

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

Square 1.

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

Add \frac{3}{2} to 1.

$$x^{2}+2x+1=\frac{5}{2}$$

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

$$\left(x+1\right)^{2}=\frac{5}{2}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+1\right)^{2}}=\sqrt{\frac{5}{2}}$$

Simplify.

$$x+1=\frac{\sqrt{10}}{2} x+1=-\frac{\sqrt{10}}{2}$$

Subtract 1 from both sides of the equation.

$$x=\frac{\sqrt{10}}{2}-1 x=-\frac{\sqrt{10}}{2}-1$$