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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(+5y^2-4y+35y-28)=0

We get rid of parentheses

5y^2-4y+35y-28=0

We add all the numbers together, and all the variables

5y^2+31y-28=0

a = 5; b = 31; c = -28;
Δ = b2-4ac
Δ = 312-4·5·(-28)
Δ = 1521
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1521}=39$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(31)-39}{2*5}=\frac{-70}{10}
=-7
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(31)+39}{2*5}=\frac{8}{10}
=4/5
$