Solve for y 3y^2+12y+4=y

$3 y^{2} + 12 y + 4 = y$

Step 1

Move all terms containing $y$ to the left side of the equation.

Subtract $y$ from both sides of the equation.

$3 y^{2} + 12 y + 4 - y = 0$

Subtract $y$ from $12 y$ .

$3 y^{2} + 11 y + 4 = 0$

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 $a = 3$ , $b = 11$ , and $c = 4$ into the quadratic formula and solve for $y$ .

$\frac{- 11 \pm \sqrt{11^{2} - 4 \cdot (3 \cdot 4)}}{2 \cdot 3}$

Step 4

Simplify.

Simplify the numerator.

Raise $11$ to the power of $2$ .

$y = \frac{- 11 \pm \sqrt{121 - 4 \cdot 3 \cdot 4}}{2 \cdot 3}$

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

Multiply $- 4$ by $3$ .

$y = \frac{- 11 \pm \sqrt{121 - 12 \cdot 4}}{2 \cdot 3}$

Multiply $- 12$ by $4$ .

$y = \frac{- 11 \pm \sqrt{121 - 48}}{2 \cdot 3}$

Subtract $48$ from $121$ .

$y = \frac{- 11 \pm \sqrt{73}}{2 \cdot 3}$

Multiply $2$ by $3$ .

$y = \frac{- 11 \pm \sqrt{73}}{6}$

Step 5

The final answer is the combination of both solutions.

$y = - \frac{11 - \sqrt{73}}{6}, - \frac{11 + \sqrt{73}}{6}$

Step 6

The result can be shown in multiple forms.

Exact Form:

$y = - \frac{11 - \sqrt{73}}{6}, - \frac{11 + \sqrt{73}}{6}$

Decimal Form:

$y = - 0.40933270 \dots, - 3.25733395 \dots$