# How do you solve y^2-90=13y?

Jul 6, 2017

I used the quadratic formula, and found:

$y = 18 \mathmr{and} y = - 5$

#### Explanation:

We have:

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

Rearrange into standard form:

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

Standard form has the powers of the pronumeral in decreasing order on the left and $0$ on the right. It can be written as:

$a {y}^{2} + b y + c = 0$, where $a$, $b$ and $c$ are the coefficients of the powers of the pronumeral.

In this case, $a = 1$, $b = - 13$ and $c = - 90$.

We often use $x$ as the pronumeral, but we treat $y$ just the same way. The quadratic formula in terms of $y$ is just:

$y = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

Substituting in the values from the equation:

$y = \frac{13 \pm \sqrt{{\left(- 13\right)}^{2} - 4 \left(1\right) \left(- 90\right)}}{2 \left(1\right)}$

$y = \frac{13 \pm \sqrt{169 + 360}}{2} = \frac{13 \pm 23}{2}$

Therefore:

$y = 18 \mathmr{and} y = - 5$

This quadratic could also be solved by factorising.

Jul 6, 2017

$y = 18 \mathmr{and} - 5$

#### Explanation:

Write as:$\text{ } {y}^{2} - 13 y - 90 = 0$

This is no different to a quadratic in $x$ in format. The only difference is that it is in $y$. So the whole thing is:

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

A quadratic in $x$ where $a {x}^{2} > 0$ gives $\cup$
A quadratic in $y$ where $a {y}^{2} > 0$ gives $\subset$
So it is rotating $\cup \textcolor{w h i t e}{.} {90}^{o}$ clockwise.

Lets check the whole number factors of 90 to see if any of them give a difference of 13

$1 \times 90 = 90$
$2 \times 45 = 90$
$3 \times 30 = 90$
4xx?=90larr" ? will not be a whole number factor"
$\textcolor{p u r p \le}{5 \times 18 = 90 \leftarrow \text{ difference is 13}}$

The 90 is negative so the two numbers are of opposite sign.
The 13 is negative so the greater of the two is negative giving:

$\left(y - 18\right) \left(y + 5\right) = 0$

Condition 1 : $\text{ "y-18=0 " "=>" } y = + 18$
Condition 2: $\text{ "y+color(white)(1)5=0" "=>" } y = - \textcolor{w h i t e}{1} 5$

Thus $y = 18 \mathmr{and} - 5$