(5+y)(4y-3)=0

Solution for (5+y)(4y-3)=0 equation:

(5+y)(4y-3)=0

We add all the numbers together, and all the variables

(y+5)(4y-3)=0

We multiply parentheses ..

(+4y^2-3y+20y-15)=0

We get rid of parentheses

4y^2-3y+20y-15=0

We add all the numbers together, and all the variables

4y^2+17y-15=0

a = 4; b = 17; c = -15;
Δ = b2-4ac
Δ = 172-4·4·(-15)
Δ = 529
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{529}=23$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(17)-23}{2*4}=\frac{-40}{8}
=-5
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(17)+23}{2*4}=\frac{6}{8}
=3/4
$