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

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

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

We move all terms to the left:

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

We multiply parentheses ..

(+4y^2-6y-10y+15)=0

We get rid of parentheses

4y^2-6y-10y+15=0

We add all the numbers together, and all the variables

4y^2-16y+15=0

a = 4; b = -16; c = +15;
Δ = b2-4ac
Δ = -162-4·4·15
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-4}{2*4}=\frac{12}{8}
=1+1/2
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+4}{2*4}=\frac{20}{8}
=2+1/2
$