(4y-9)(5+y)=0

Solution for (4y-9)(5+y)=0 equation:

(4y-9)(5+y)=0

We add all the numbers together, and all the variables

(4y-9)(y+5)=0

We multiply parentheses ..

(+4y^2+20y-9y-45)=0

We get rid of parentheses

4y^2+20y-9y-45=0

We add all the numbers together, and all the variables

4y^2+11y-45=0

a = 4; b = 11; c = -45;
Δ = b2-4ac
Δ = 112-4·4·(-45)
Δ = 841
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{841}=29$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-29}{2*4}=\frac{-40}{8}
=-5
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+29}{2*4}=\frac{18}{8}
=2+1/4
$