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

Solution for (7+y)(3y-4)=0 equation:

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

We add all the numbers together, and all the variables

(y+7)(3y-4)=0

We multiply parentheses ..

(+3y^2-4y+21y-28)=0

We get rid of parentheses

3y^2-4y+21y-28=0

We add all the numbers together, and all the variables

3y^2+17y-28=0

a = 3; b = 17; c = -28;
Δ = b2-4ac
Δ = 172-4·3·(-28)
Δ = 625
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{625}=25$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(17)-25}{2*3}=\frac{-42}{6}
=-7
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(17)+25}{2*3}=\frac{8}{6}
=1+1/3
$