(2y-1)(y-1)=(y-3)(y+13)

Solution for (2y-1)(y-1)=(y-3)(y+13) equation:

(2y-1)(y-1)=(y-3)(y+13)

We move all terms to the left:

(2y-1)(y-1)-((y-3)(y+13))=0

We multiply parentheses ..

(+2y^2-2y-1y+1)-((y-3)(y+13))=0


We calculate terms in parentheses: -((y-3)(y+13)), so:

(y-3)(y+13)

We multiply parentheses ..

(+y^2+13y-3y-39)

We get rid of parentheses

y^2+13y-3y-39

We add all the numbers together, and all the variables

y^2+10y-39

Back to the equation:

-(y^2+10y-39)


We get rid of parentheses

2y^2-y^2-2y-1y-10y+1+39=0

We add all the numbers together, and all the variables

y^2-13y+40=0

a = 1; b = -13; c = +40;
Δ = b2-4ac
Δ = -132-4·1·40
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-13)-3}{2*1}=\frac{10}{2}
=5
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+3}{2*1}=\frac{16}{2}
=8
$