(3y-5)(2y-1)=10

Solution for (3y-5)(2y-1)=10 equation:

(3y-5)(2y-1)=10

We move all terms to the left:

(3y-5)(2y-1)-(10)=0

We multiply parentheses ..

(+6y^2-3y-10y+5)-10=0

We get rid of parentheses

6y^2-3y-10y+5-10=0

We add all the numbers together, and all the variables

6y^2-13y-5=0

a = 6; b = -13; c = -5;
Δ = b2-4ac
Δ = -132-4·6·(-5)
Δ = 289
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{289}=17$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-13)-17}{2*6}=\frac{-4}{12}
=-1/3
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+17}{2*6}=\frac{30}{12}
=2+1/2
$