(7+y)(3y-2)=0

Solution for (7+y)(3y-2)=0 equation:

(7+y)(3y-2)=0

We add all the numbers together, and all the variables

(y+7)(3y-2)=0

We multiply parentheses ..

(+3y^2-2y+21y-14)=0

We get rid of parentheses

3y^2-2y+21y-14=0

We add all the numbers together, and all the variables

3y^2+19y-14=0

a = 3; b = 19; c = -14;
Δ = b2-4ac
Δ = 192-4·3·(-14)
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{-(19)-23}{2*3}=\frac{-42}{6}
=-7
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+23}{2*3}=\frac{4}{6}
=2/3
$