Solve Using the Quadratic Formula -3y^2=6y-10

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

Step 1

Move all the expressions to the left side of the equation.

Subtract $6 y$ from both sides of the equation.

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

Add $10$ to both sides of the equation.

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

Step 2

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 3

Substitute the values $a = - 3$ , $b = - 6$ , and $c = 10$ into the quadratic formula and solve for $y$ .

$\frac{6 \pm \sqrt{{(- 6)}^{2} - 4 \cdot (- 3 \cdot 10)}}{2 \cdot - 3}$

Step 4

Simplify.

Simplify the numerator.

Raise $- 6$ to the power of $2$ .

$y = \frac{6 \pm \sqrt{36 - 4 \cdot - 3 \cdot 10}}{2 \cdot - 3}$

Multiply $- 4 \cdot - 3 \cdot 10$ .

Multiply $- 4$ by $- 3$ .

$y = \frac{6 \pm \sqrt{36 + 12 \cdot 10}}{2 \cdot - 3}$

Multiply $12$ by $10$ .

$y = \frac{6 \pm \sqrt{36 + 120}}{2 \cdot - 3}$

Add $36$ and $120$ .

$y = \frac{6 \pm \sqrt{156}}{2 \cdot - 3}$

Rewrite $156$ as $2^{2} \cdot 39$ .

Factor $4$ out of $156$ .

$y = \frac{6 \pm \sqrt{4 (39)}}{2 \cdot - 3}$

Rewrite $4$ as $2^{2}$ .

$y = \frac{6 \pm \sqrt{2^{2} \cdot 39}}{2 \cdot - 3}$

Pull terms out from under the radical.

$y = \frac{6 \pm 2 \sqrt{39}}{2 \cdot - 3}$

Multiply $2$ by $- 3$ .

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

Simplify $\frac{6 \pm 2 \sqrt{39}}{- 6}$ .

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

Move the negative in front of the fraction.

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

Step 5

The result can be shown in multiple forms.

Exact Form:

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

Decimal Form:

$y = - 3.08166599 \dots, 1.08166599 \dots$