(2y-5)(y+1)=(3y-7)(y+1)

Solution for (2y-5)(y+1)=(3y-7)(y+1) equation:

(2y-5)(y+1)=(3y-7)(y+1)

We move all terms to the left:

(2y-5)(y+1)-((3y-7)(y+1))=0

We multiply parentheses ..

(+2y^2+2y-5y-5)-((3y-7)(y+1))=0


We calculate terms in parentheses: -((3y-7)(y+1)), so:

(3y-7)(y+1)

We multiply parentheses ..

(+3y^2+3y-7y-7)

We get rid of parentheses

3y^2+3y-7y-7

We add all the numbers together, and all the variables

3y^2-4y-7

Back to the equation:

-(3y^2-4y-7)


We get rid of parentheses

2y^2-3y^2+2y-5y+4y-5+7=0

We add all the numbers together, and all the variables

-1y^2+y+2=0

a = -1; b = 1; c = +2;
Δ = b2-4ac
Δ = 12-4·(-1)·2
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{-(1)-3}{2*-1}=\frac{-4}{-2}
=+2
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+3}{2*-1}=\frac{2}{-2}
=-1
$