(5+y)(4y-7)=0

Solution for (5+y)(4y-7)=0 equation:

(5+y)(4y-7)=0

We add all the numbers together, and all the variables

(y+5)(4y-7)=0

We multiply parentheses ..

(+4y^2-7y+20y-35)=0

We get rid of parentheses

4y^2-7y+20y-35=0

We add all the numbers together, and all the variables

4y^2+13y-35=0

a = 4; b = 13; c = -35;
Δ = b2-4ac
Δ = 132-4·4·(-35)
Δ = 729
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{729}=27$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-27}{2*4}=\frac{-40}{8}
=-5
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+27}{2*4}=\frac{14}{8}
=1+3/4
$