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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(+4y^2+36y-5y-45)=0

We get rid of parentheses

4y^2+36y-5y-45=0

We add all the numbers together, and all the variables

4y^2+31y-45=0

a = 4; b = 31; c = -45;
Δ = b2-4ac
Δ = 312-4·4·(-45)
Δ = 1681
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1681}=41$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(31)-41}{2*4}=\frac{-72}{8}
=-9
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(31)+41}{2*4}=\frac{10}{8}
=1+1/4
$