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

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

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

We move all terms to the left:

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

We multiply parentheses ..

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


We calculate terms in parentheses: -(y(y+5)), so:

y(y+5)

We multiply parentheses

y^2+5y

Back to the equation:

-(y^2+5y)


We get rid of parentheses

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

We add all the numbers together, and all the variables

y^2-10y-3=0

a = 1; b = -10; c = -3;
Δ = b2-4ac
Δ = -102-4·1·(-3)
Δ = 112
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:

$\sqrt{\Delta}=\sqrt{112}=\sqrt{16*7}=\sqrt{16}*\sqrt{7}=4\sqrt{7}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-4\sqrt{7}}{2*1}=\frac{10-4\sqrt{7}}{2}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+4\sqrt{7}}{2*1}=\frac{10+4\sqrt{7}}{2}
$