Solve y^2-8y=-11

$$y ^ { 2 } - 8 y = - 11$$

Solve for y

$$y=\sqrt{5}+4\approx 6.236067977$$

$$y=4-\sqrt{5}\approx 1.763932023$$

Steps Using the Quadratic Formula

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.

$$y^{2}-8y=-11$$

Add 11 to both sides of the equation.

$$y^{2}-8y-\left(-11\right)=-11-\left(-11\right)$$

Subtracting -11 from itself leaves 0.

$$y^{2}-8y-\left(-11\right)=0$$

Subtract -11 from 0.

$$y^{2}-8y+11=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -8 for b, and 11 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\times 11}}{2}$$

Square -8.

$$y=\frac{-\left(-8\right)±\sqrt{64-4\times 11}}{2}$$

Multiply -4 times 11.

$$y=\frac{-\left(-8\right)±\sqrt{64-44}}{2}$$

Add 64 to -44.

$$y=\frac{-\left(-8\right)±\sqrt{20}}{2}$$

Take the square root of 20.

$$y=\frac{-\left(-8\right)±2\sqrt{5}}{2}$$

The opposite of -8 is 8.

$$y=\frac{8±2\sqrt{5}}{2}$$

Now solve the equation y=\frac{8±2\sqrt{5}}{2} when ± is plus. Add 8 to 2\sqrt{5}.

$$y=\frac{2\sqrt{5}+8}{2}$$

Divide 8+2\sqrt{5} by 2.

$$y=\sqrt{5}+4$$

Now solve the equation y=\frac{8±2\sqrt{5}}{2} when ± is minus. Subtract 2\sqrt{5} from 8.

$$y=\frac{8-2\sqrt{5}}{2}$$

Divide 8-2\sqrt{5} by 2.

$$y=4-\sqrt{5}$$

The equation is now solved.

$$y=\sqrt{5}+4 y=4-\sqrt{5}$$

Steps for Completing the Square

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$y^{2}-8y=-11$$

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.

$$y^{2}-8y+\left(-4\right)^{2}=-11+\left(-4\right)^{2}$$

Square -4.

$$y^{2}-8y+16=-11+16$$

Add -11 to 16.

$$y^{2}-8y+16=5$$

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}.

$$\left(y-4\right)^{2}=5$$

Take the square root of both sides of the equation.

$$\sqrt{\left(y-4\right)^{2}}=\sqrt{5}$$

Simplify.

$$y-4=\sqrt{5} y-4=-\sqrt{5}$$

Add 4 to both sides of the equation.

$$y=\sqrt{5}+4 y=4-\sqrt{5}$$