Solve y^2-90=13y

$$y ^ { 2 } - 90 = 13 y$$

Solve for y

$$y=-5$$

$$y=18$$

Steps Using Factoring

Subtract 13y from both sides.

$$y^{2}-90-13y=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$y^{2}-13y-90=0$$

To solve the equation, factor y^{2}-13y-90 using formula y^{2}+\left(a+b\right)y+ab=\left(y+a\right)\left(y+b\right). To find a and b, set up a system to be solved.

$$a+b=-13 ab=-90$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -90.

$$1,-90 2,-45 3,-30 5,-18 6,-15 9,-10$$

Calculate the sum for each pair.

$$1-90=-89 2-45=-43 3-30=-27 5-18=-13 6-15=-9 9-10=-1$$

The solution is the pair that gives sum -13.

$$a=-18 b=5$$

Rewrite factored expression \left(y+a\right)\left(y+b\right) using the obtained values.

$$\left(y-18\right)\left(y+5\right)$$

To find equation solutions, solve y-18=0 and y+5=0.

$$y=18 y=-5$$

Steps Using Factoring By Grouping

Subtract 13y from both sides.

$$y^{2}-90-13y=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$y^{2}-13y-90=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as y^{2}+ay+by-90. To find a and b, set up a system to be solved.

$$a+b=-13 ab=1\left(-90\right)=-90$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -90.

$$1,-90 2,-45 3,-30 5,-18 6,-15 9,-10$$

Calculate the sum for each pair.

$$1-90=-89 2-45=-43 3-30=-27 5-18=-13 6-15=-9 9-10=-1$$

The solution is the pair that gives sum -13.

$$a=-18 b=5$$

Rewrite y^{2}-13y-90 as \left(y^{2}-18y\right)+\left(5y-90\right).

$$\left(y^{2}-18y\right)+\left(5y-90\right)$$

Factor out y in the first and 5 in the second group.

$$y\left(y-18\right)+5\left(y-18\right)$$

Factor out common term y-18 by using distributive property.

$$\left(y-18\right)\left(y+5\right)$$

To find equation solutions, solve y-18=0 and y+5=0.

$$y=18 y=-5$$

Steps Using the Quadratic Formula

Subtract 13y from both sides.

$$y^{2}-90-13y=0$$

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.

$$y^{2}-13y-90=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -13 for b, and -90 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\left(-13\right)±\sqrt{\left(-13\right)^{2}-4\left(-90\right)}}{2}$$

Square -13.

$$y=\frac{-\left(-13\right)±\sqrt{169-4\left(-90\right)}}{2}$$

Multiply -4 times -90.

$$y=\frac{-\left(-13\right)±\sqrt{169+360}}{2}$$

Add 169 to 360.

$$y=\frac{-\left(-13\right)±\sqrt{529}}{2}$$

Take the square root of 529.

$$y=\frac{-\left(-13\right)±23}{2}$$

The opposite of -13 is 13.

$$y=\frac{13±23}{2}$$

Now solve the equation y=\frac{13±23}{2} when ± is plus. Add 13 to 23.

$$y=\frac{36}{2}$$

Divide 36 by 2.

$$y=18$$

Now solve the equation y=\frac{13±23}{2} when ± is minus. Subtract 23 from 13.

$$y=-\frac{10}{2}$$

Divide -10 by 2.

$$y=-5$$

The equation is now solved.

$$y=18 y=-5$$

Steps for Completing the Square

Subtract 13y from both sides.

$$y^{2}-90-13y=0$$

Add 90 to both sides. Anything plus zero gives itself.

$$y^{2}-13y=90$$

Divide -13, the coefficient of the x term, by 2 to get -\frac{13}{2}. Then add the square of -\frac{13}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$y^{2}-13y+\left(-\frac{13}{2}\right)^{2}=90+\left(-\frac{13}{2}\right)^{2}$$

Square -\frac{13}{2} by squaring both the numerator and the denominator of the fraction.

$$y^{2}-13y+\frac{169}{4}=90+\frac{169}{4}$$

Add 90 to \frac{169}{4}.

$$y^{2}-13y+\frac{169}{4}=\frac{529}{4}$$

Factor y^{2}-13y+\frac{169}{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}.

$$\left(y-\frac{13}{2}\right)^{2}=\frac{529}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(y-\frac{13}{2}\right)^{2}}=\sqrt{\frac{529}{4}}$$

Simplify.

$$y-\frac{13}{2}=\frac{23}{2} y-\frac{13}{2}=-\frac{23}{2}$$

Add \frac{13}{2} to both sides of the equation.

$$y=18 y=-5$$