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

Solution for (7+y)(3y-5)=0 equation:

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

We add all the numbers together, and all the variables

(y+7)(3y-5)=0

We multiply parentheses ..

(+3y^2-5y+21y-35)=0

We get rid of parentheses

3y^2-5y+21y-35=0

We add all the numbers together, and all the variables

3y^2+16y-35=0

a = 3; b = 16; c = -35;
Δ = b2-4ac
Δ = 162-4·3·(-35)
Δ = 676
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{676}=26$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-26}{2*3}=\frac{-42}{6}
=-7
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+26}{2*3}=\frac{10}{6}
=1+2/3
$