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

Solution for (1+y)(4y-9)=0 equation:

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

We add all the numbers together, and all the variables

(y+1)(4y-9)=0

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

4y^2-5y-9=0

a = 4; b = -5; c = -9;
Δ = b2-4ac
Δ = -52-4·4·(-9)
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-13}{2*4}=\frac{-8}{8}
=-1
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+13}{2*4}=\frac{18}{8}
=2+1/4
$