Solve 16+y^2-8y-12=0

$$16 + y ^ { 2 } - 8 y - 12 = 0$$

Solve for y

$$y=2\sqrt{3}+4\approx 7.464101615$$

$$y=4-2\sqrt{3}\approx 0.535898385$$

Steps Using the Quadratic Formula

Subtract 12 from 16 to get 4.

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

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -8 for b, and 4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\times 4}}{2}$$

Square -8.

$$y=\frac{-\left(-8\right)±\sqrt{64-4\times 4}}{2}$$

Multiply -4 times 4.

$$y=\frac{-\left(-8\right)±\sqrt{64-16}}{2}$$

Add 64 to -16.

$$y=\frac{-\left(-8\right)±\sqrt{48}}{2}$$

Take the square root of 48.

$$y=\frac{-\left(-8\right)±4\sqrt{3}}{2}$$

The opposite of -8 is 8.

$$y=\frac{8±4\sqrt{3}}{2}$$

Now solve the equation y=\frac{8±4\sqrt{3}}{2} when ± is plus. Add 8 to 4\sqrt{3}.

$$y=\frac{4\sqrt{3}+8}{2}$$

Divide 8+4\sqrt{3} by 2.

$$y=2\sqrt{3}+4$$

Now solve the equation y=\frac{8±4\sqrt{3}}{2} when ± is minus. Subtract 4\sqrt{3} from 8.

$$y=\frac{8-4\sqrt{3}}{2}$$

Divide 8-4\sqrt{3} by 2.

$$y=4-2\sqrt{3}$$

The equation is now solved.

$$y=2\sqrt{3}+4 y=4-2\sqrt{3}$$

Steps for Completing the Square

Subtract 12 from 16 to get 4.

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

Subtract 4 from both sides. Anything subtracted from zero gives its negation.

$$y^{2}-8y=-4$$

Divide -8, the coefficient of the x term, by 2 to get -4. Then add the square of -4 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$y^{2}-8y+\left(-4\right)^{2}=-4+\left(-4\right)^{2}$$

Square -4.

$$y^{2}-8y+16=-4+16$$

Add -4 to 16.

$$y^{2}-8y+16=12$$

Factor y^{2}-8y+16. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$y-4=2\sqrt{3} y-4=-2\sqrt{3}$$

Add 4 to both sides of the equation.

$$y=2\sqrt{3}+4 y=4-2\sqrt{3}$$