(2y+9)(3-y)=0

Solution for (2y+9)(3-y)=0 equation:

(2y+9)(3-y)=0

We add all the numbers together, and all the variables

(2y+9)(-1y+3)=0

We multiply parentheses ..

(-2y^2+6y-9y+27)=0

We get rid of parentheses

-2y^2+6y-9y+27=0

We add all the numbers together, and all the variables

-2y^2-3y+27=0

a = -2; b = -3; c = +27;
Δ = b2-4ac
Δ = -32-4·(-2)·27
Δ = 225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{225}=15$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-15}{2*-2}=\frac{-12}{-4}
=+3
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+15}{2*-2}=\frac{18}{-4}
=-4+1/2
$