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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(+4y^2+4y-4y-4)-((10y-1)(-2.5))=0


We calculate terms in parentheses: -((10y-1)(-2.5)), so:

(10y-1)(-2.5)

We multiply parentheses ..

(-25y+2.5)

We get rid of parentheses

-25y+2.5

Back to the equation:

-(-25y+2.5)


We get rid of parentheses

4y^2+4y-4y+25y-4-2.5=0

We add all the numbers together, and all the variables

4y^2+25y-6.5=0

a = 4; b = 25; c = -6.5;
Δ = b2-4ac
Δ = 252-4·4·(-6.5)
Δ = 729
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{729}=27$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(25)-27}{2*4}=\frac{-52}{8}
=-6+1/2
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(25)+27}{2*4}=\frac{2}{8}
=1/4
$