(2y-3)5y=10(y-1)

Solution for (2y-3)5y=10(y-1) equation:

(2y-3)5y=10(y-1)

We move all terms to the left:

(2y-3)5y-(10(y-1))=0

We multiply parentheses

10y^2-15y-(10(y-1))=0


We calculate terms in parentheses: -(10(y-1)), so:

10(y-1)

We multiply parentheses

10y-10

Back to the equation:

-(10y-10)


We get rid of parentheses

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

We add all the numbers together, and all the variables

10y^2-25y+10=0

a = 10; b = -25; c = +10;
Δ = b2-4ac
Δ = -252-4·10·10
Δ = 225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{225}=15$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-25)-15}{2*10}=\frac{10}{20}
=1/2
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-25)+15}{2*10}=\frac{40}{20}
=2
$