Solve y^2+4y+8y-12=0

$$y ^ { 2 } + 4 y + 8 y - 12 = 0$$

Solve for y

$$y=4\sqrt{3}-6\approx 0.92820323$$

$$y=-4\sqrt{3}-6\approx -12.92820323$$

Steps Using the Quadratic Formula

Combine 4y and 8y to get 12y.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 12 for b, and -12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-12±\sqrt{12^{2}-4\left(-12\right)}}{2}$$

Square 12.

$$y=\frac{-12±\sqrt{144-4\left(-12\right)}}{2}$$

Multiply -4 times -12.

$$y=\frac{-12±\sqrt{144+48}}{2}$$

Add 144 to 48.

$$y=\frac{-12±\sqrt{192}}{2}$$

Take the square root of 192.

$$y=\frac{-12±8\sqrt{3}}{2}$$

Now solve the equation y=\frac{-12±8\sqrt{3}}{2} when ± is plus. Add -12 to 8\sqrt{3}.

$$y=\frac{8\sqrt{3}-12}{2}$$

Divide -12+8\sqrt{3} by 2.

$$y=4\sqrt{3}-6$$

Now solve the equation y=\frac{-12±8\sqrt{3}}{2} when ± is minus. Subtract 8\sqrt{3} from -12.

$$y=\frac{-8\sqrt{3}-12}{2}$$

Divide -12-8\sqrt{3} by 2.

$$y=-4\sqrt{3}-6$$

The equation is now solved.

$$y=4\sqrt{3}-6 y=-4\sqrt{3}-6$$

Steps for Completing the Square

Combine 4y and 8y to get 12y.

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

Add 12 to both sides. Anything plus zero gives itself.

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

Divide 12, the coefficient of the x term, by 2 to get 6. Then add the square of 6 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$y^{2}+12y+6^{2}=12+6^{2}$$

Square 6.

$$y^{2}+12y+36=12+36$$

Add 12 to 36.

$$y^{2}+12y+36=48$$

Factor y^{2}+12y+36. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(y+6\right)^{2}=48$$

Take the square root of both sides of the equation.

$$\sqrt{\left(y+6\right)^{2}}=\sqrt{48}$$

Simplify.

$$y+6=4\sqrt{3} y+6=-4\sqrt{3}$$

Subtract 6 from both sides of the equation.

$$y=4\sqrt{3}-6 y=-4\sqrt{3}-6$$