(4y+9)(y-2)=0

Solution for (4y+9)(y-2)=0 equation:

(4y+9)(y-2)=0

We multiply parentheses ..

(+4y^2-8y+9y-18)=0

We get rid of parentheses

4y^2-8y+9y-18=0

We add all the numbers together, and all the variables

4y^2+y-18=0

a = 4; b = 1; c = -18;
Δ = b2-4ac
Δ = 12-4·4·(-18)
Δ = 289
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{289}=17$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-17}{2*4}=\frac{-18}{8}
=-2+1/4
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+17}{2*4}=\frac{16}{8}
=2
$