# How do you solve y^2 - 5y = -4?

Mar 10, 2018

See a solution process below:

#### Explanation:

First, add $\textcolor{red}{4}$ to each side of the equation to put it into standard quadratic form:

${y}^{2} - 5 y + \textcolor{red}{4} = - 4 + \textcolor{red}{4}$

${y}^{2} - 5 y + 4 = 0$

Next, factor as:

$\left(y - 1\right) \left(y - 4\right) = 0$

Now, solve each term on the left for $0$:

Solution 1:

$y - 1 = 0$

$y - 1 + \textcolor{red}{1} = 0 + \textcolor{red}{1}$

$y - 0 = 1$

$y = 1$

Solution 2:

$y - 4 = 0$

$y - 4 + \textcolor{red}{4} = 0 + \textcolor{red}{4}$

$y - 0 = 4$

$y = 4$

The Solution Set Is: $y = \left\{1 , 4\right\}$