-5y(3y-2)=-2y+4y-7

Solution for -5y(3y-2)=-2y+4y-7 equation:

-5y(3y-2)=-2y+4y-7

We move all terms to the left:

-5y(3y-2)-(-2y+4y-7)=0

We add all the numbers together, and all the variables

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

We multiply parentheses

-15y^2+10y-(2y-7)=0

We get rid of parentheses

-15y^2+10y-2y+7=0

We add all the numbers together, and all the variables

-15y^2+8y+7=0

a = -15; b = 8; c = +7;
Δ = b2-4ac
Δ = 82-4·(-15)·7
Δ = 484
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{484}=22$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-22}{2*-15}=\frac{-30}{-30}
=1
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+22}{2*-15}=\frac{14}{-30}
=-7/15
$