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

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

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

We move all terms to the left:

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

We multiply parentheses ..

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


We calculate terms in parentheses: -(y(y-2)), so:

y(y-2)

We multiply parentheses

y^2-2y

Back to the equation:

-(y^2-2y)


We get rid of parentheses

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

We add all the numbers together, and all the variables

2y^2+y-10=0

a = 2; b = 1; c = -10;
Δ = b2-4ac
Δ = 12-4·2·(-10)
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-9}{2*2}=\frac{-10}{4}
=-2+1/2
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+9}{2*2}=\frac{8}{4}
=2
$