(3y+5)(2y-7)=27

Solution for (3y+5)(2y-7)=27 equation:

(3y+5)(2y-7)=27

We move all terms to the left:

(3y+5)(2y-7)-(27)=0

We multiply parentheses ..

(+6y^2-21y+10y-35)-27=0

We get rid of parentheses

6y^2-21y+10y-35-27=0

We add all the numbers together, and all the variables

6y^2-11y-62=0

a = 6; b = -11; c = -62;
Δ = b2-4ac
Δ = -112-4·6·(-62)
Δ = 1609
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11)-\sqrt{1609}}{2*6}=\frac{11-\sqrt{1609}}{12}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+\sqrt{1609}}{2*6}=\frac{11+\sqrt{1609}}{12}
$