(3y+4)(10-y)=2

Solution for (3y+4)(10-y)=2 equation:

(3y+4)(10-y)=2

We move all terms to the left:

(3y+4)(10-y)-(2)=0

We add all the numbers together, and all the variables

(3y+4)(-1y+10)-2=0

We multiply parentheses ..

(-3y^2+30y-4y+40)-2=0

We get rid of parentheses

-3y^2+30y-4y+40-2=0

We add all the numbers together, and all the variables

-3y^2+26y+38=0

a = -3; b = 26; c = +38;
Δ = b2-4ac
Δ = 262-4·(-3)·38
Δ = 1132
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1132}=\sqrt{4*283}=\sqrt{4}*\sqrt{283}=2\sqrt{283}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(26)-2\sqrt{283}}{2*-3}=\frac{-26-2\sqrt{283}}{-6}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(26)+2\sqrt{283}}{2*-3}=\frac{-26+2\sqrt{283}}{-6}
$