(2.5y-2)(2y+2)=(10y-1)(0.5y-3)

Solution for (2.5y-2)(2y+2)=(10y-1)(0.5y-3) equation:

(2.5y-2)(2y+2)=(10y-1)(0.5y-3)

We move all terms to the left:

(2.5y-2)(2y+2)-((10y-1)(0.5y-3))=0

We multiply parentheses ..

(+4y^2+4y-4y-4)-((10y-1)(0.5y-3))=0


We calculate terms in parentheses: -((10y-1)(0.5y-3)), so:

(10y-1)(0.5y-3)

We multiply parentheses ..

(+0y^2-30y+0y+3)

We get rid of parentheses

0y^2-30y+0y+3

We add all the numbers together, and all the variables

y^2-29y+3

Back to the equation:

-(y^2-29y+3)


We get rid of parentheses

4y^2-y^2+4y-4y+29y-4-3=0

We add all the numbers together, and all the variables

3y^2+29y-7=0

a = 3; b = 29; c = -7;
Δ = b2-4ac
Δ = 292-4·3·(-7)
Δ = 925
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:

$\sqrt{\Delta}=\sqrt{925}=\sqrt{25*37}=\sqrt{25}*\sqrt{37}=5\sqrt{37}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(29)-5\sqrt{37}}{2*3}=\frac{-29-5\sqrt{37}}{6}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(29)+5\sqrt{37}}{2*3}=\frac{-29+5\sqrt{37}}{6}
$