y^2+(-5y)+4=0

Solution for y^2+(-5y)+4=0 equation:

y^2+(-5y)+4=0

We get rid of parentheses

y^2-5y+4=0

a = 1; b = -5; c = +4;
Δ = b2-4ac
Δ = -52-4·1·4
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-3}{2*1}=\frac{2}{2}
=1
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+3}{2*1}=\frac{8}{2}
=4
$