Solve y^2+289/4=-17y

$$y ^ { 2 } + \frac { 289 } { 4 } = - 17 y$$

Solve for y

$$y = -\frac{17}{2} = -8\frac{1}{2} = -8.5$$

Steps Using the Quadratic Formula

Add 17y to both sides.

$$y^{2}+\frac{289}{4}+17y=0$$

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.

$$y^{2}+17y+\frac{289}{4}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 17 for b, and \frac{289}{4} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-17±\sqrt{17^{2}-4\times \frac{289}{4}}}{2}$$

Square 17.

$$y=\frac{-17±\sqrt{289-4\times \frac{289}{4}}}{2}$$

Multiply -4 times \frac{289}{4}.

$$y=\frac{-17±\sqrt{289-289}}{2}$$

Add 289 to -289.

$$y=\frac{-17±\sqrt{0}}{2}$$

Take the square root of 0.

$$y=-\frac{17}{2}$$

Steps for Completing the Square

Add 17y to both sides.

$$y^{2}+\frac{289}{4}+17y=0$$

Subtract \frac{289}{4} from both sides. Anything subtracted from zero gives its negation.

$$y^{2}+17y=-\frac{289}{4}$$

Divide 17, the coefficient of the x term, by 2 to get \frac{17}{2}. Then add the square of \frac{17}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$y^{2}+17y+\left(\frac{17}{2}\right)^{2}=-\frac{289}{4}+\left(\frac{17}{2}\right)^{2}$$

Square \frac{17}{2} by squaring both the numerator and the denominator of the fraction.

$$y^{2}+17y+\frac{289}{4}=\frac{-289+289}{4}$$

Add -\frac{289}{4} to \frac{289}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$y^{2}+17y+\frac{289}{4}=0$$

Factor y^{2}+17y+\frac{289}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(y+\frac{17}{2}\right)^{2}=0$$

Take the square root of both sides of the equation.

$$\sqrt{\left(y+\frac{17}{2}\right)^{2}}=\sqrt{0}$$

Simplify.

$$y+\frac{17}{2}=0 y+\frac{17}{2}=0$$

Subtract \frac{17}{2} from both sides of the equation.

$$y=-\frac{17}{2} y=-\frac{17}{2}$$

The equation is now solved. Solutions are the same.

$$y=-\frac{17}{2}$$