Solve 0=-3x^2+6x+2

$$0 = - 3 x ^ { 2 } + 6 x + 2$$

Solve for x

$$x=\frac{\sqrt{15}}{3}+1\approx 2.290994449$$

$$x=-\frac{\sqrt{15}}{3}+1\approx -0.290994449$$

Steps Using the Quadratic Formula

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

$$-3x^{2}+6x+2=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -3 for a, 6 for b, and 2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-6±\sqrt{6^{2}-4\left(-3\right)\times 2}}{2\left(-3\right)}$$

Square 6.

$$x=\frac{-6±\sqrt{36-4\left(-3\right)\times 2}}{2\left(-3\right)}$$

Multiply -4 times -3.

$$x=\frac{-6±\sqrt{36+12\times 2}}{2\left(-3\right)}$$

Multiply 12 times 2.

$$x=\frac{-6±\sqrt{36+24}}{2\left(-3\right)}$$

Add 36 to 24.

$$x=\frac{-6±\sqrt{60}}{2\left(-3\right)}$$

Take the square root of 60.

$$x=\frac{-6±2\sqrt{15}}{2\left(-3\right)}$$

Multiply 2 times -3.

$$x=\frac{-6±2\sqrt{15}}{-6}$$

Now solve the equation x=\frac{-6±2\sqrt{15}}{-6} when ± is plus. Add -6 to 2\sqrt{15}.

$$x=\frac{2\sqrt{15}-6}{-6}$$

Divide -6+2\sqrt{15} by -6.

$$x=-\frac{\sqrt{15}}{3}+1$$

Now solve the equation x=\frac{-6±2\sqrt{15}}{-6} when ± is minus. Subtract 2\sqrt{15} from -6.

$$x=\frac{-2\sqrt{15}-6}{-6}$$

Divide -6-2\sqrt{15} by -6.

$$x=\frac{\sqrt{15}}{3}+1$$

The equation is now solved.

$$x=-\frac{\sqrt{15}}{3}+1 x=\frac{\sqrt{15}}{3}+1$$

Steps for Completing the Square

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

$$-3x^{2}+6x+2=0$$

Subtract 2 from both sides. Anything subtracted from zero gives its negation.

$$-3x^{2}+6x=-2$$

Divide both sides by -3.

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

Dividing by -3 undoes the multiplication by -3.

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

Divide 6 by -3.

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

Divide -2 by -3.

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

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.

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

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

$$x^{2}-2x+1=\frac{5}{3}$$

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}.

$$\left(x-1\right)^{2}=\frac{5}{3}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-1\right)^{2}}=\sqrt{\frac{5}{3}}$$

Simplify.

$$x-1=\frac{\sqrt{15}}{3} x-1=-\frac{\sqrt{15}}{3}$$

Add 1 to both sides of the equation.

$$x=\frac{\sqrt{15}}{3}+1 x=-\frac{\sqrt{15}}{3}+1$$