Solve -3y^2=6y-10

$$- 3 y ^ { 2 } = 6 y - 10$$

Solve for y

$$y=\frac{\sqrt{39}}{3}-1\approx 1.081665999$$

$$y=-\frac{\sqrt{39}}{3}-1\approx -3.081665999$$

Steps Using the Quadratic Formula

Subtract 6y from both sides.

$$-3y^{2}-6y=-10$$

Add 10 to both sides.

$$-3y^{2}-6y+10=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -3 for a, -6 for b, and 10 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\left(-3\right)\times 10}}{2\left(-3\right)}$$

Square -6.

$$y=\frac{-\left(-6\right)±\sqrt{36-4\left(-3\right)\times 10}}{2\left(-3\right)}$$

Multiply -4 times -3.

$$y=\frac{-\left(-6\right)±\sqrt{36+12\times 10}}{2\left(-3\right)}$$

Multiply 12 times 10.

$$y=\frac{-\left(-6\right)±\sqrt{36+120}}{2\left(-3\right)}$$

Add 36 to 120.

$$y=\frac{-\left(-6\right)±\sqrt{156}}{2\left(-3\right)}$$

Take the square root of 156.

$$y=\frac{-\left(-6\right)±2\sqrt{39}}{2\left(-3\right)}$$

The opposite of -6 is 6.

$$y=\frac{6±2\sqrt{39}}{2\left(-3\right)}$$

Multiply 2 times -3.

$$y=\frac{6±2\sqrt{39}}{-6}$$

Now solve the equation y=\frac{6±2\sqrt{39}}{-6} when ± is plus. Add 6 to 2\sqrt{39}.

$$y=\frac{2\sqrt{39}+6}{-6}$$

Divide 6+2\sqrt{39} by -6.

$$y=-\frac{\sqrt{39}}{3}-1$$

Now solve the equation y=\frac{6±2\sqrt{39}}{-6} when ± is minus. Subtract 2\sqrt{39} from 6.

$$y=\frac{6-2\sqrt{39}}{-6}$$

Divide 6-2\sqrt{39} by -6.

$$y=\frac{\sqrt{39}}{3}-1$$

The equation is now solved.

$$y=-\frac{\sqrt{39}}{3}-1 y=\frac{\sqrt{39}}{3}-1$$

Steps for Completing the Square

Subtract 6y from both sides.

$$-3y^{2}-6y=-10$$

Divide both sides by -3.

$$\frac{-3y^{2}-6y}{-3}=-\frac{10}{-3}$$

Dividing by -3 undoes the multiplication by -3.

$$y^{2}+\left(-\frac{6}{-3}\right)y=-\frac{10}{-3}$$

Divide -6 by -3.

$$y^{2}+2y=-\frac{10}{-3}$$

Divide -10 by -3.

$$y^{2}+2y=\frac{10}{3}$$

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.

$$y^{2}+2y+1^{2}=\frac{10}{3}+1^{2}$$

Square 1.

$$y^{2}+2y+1=\frac{10}{3}+1$$

Add \frac{10}{3} to 1.

$$y^{2}+2y+1=\frac{13}{3}$$

Factor y^{2}+2y+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}.

$$\left(y+1\right)^{2}=\frac{13}{3}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(y+1\right)^{2}}=\sqrt{\frac{13}{3}}$$

Simplify.

$$y+1=\frac{\sqrt{39}}{3} y+1=-\frac{\sqrt{39}}{3}$$

Subtract 1 from both sides of the equation.

$$y=\frac{\sqrt{39}}{3}-1 y=-\frac{\sqrt{39}}{3}-1$$