2y+3(y-8)=5y-2y(y-10)

Solution for 2y+3(y-8)=5y-2y(y-10) equation:

2y+3(y-8)=5y-2y(y-10)

We move all terms to the left:

2y+3(y-8)-(5y-2y(y-10))=0

We multiply parentheses

2y+3y-(5y-2y(y-10))-24=0


We calculate terms in parentheses: -(5y-2y(y-10)), so:

5y-2y(y-10)

We multiply parentheses

-2y^2+5y+20y

We add all the numbers together, and all the variables

-2y^2+25y

Back to the equation:

-(-2y^2+25y)


We add all the numbers together, and all the variables

-(-2y^2+25y)+5y-24=0

We get rid of parentheses

2y^2-25y+5y-24=0

We add all the numbers together, and all the variables

2y^2-20y-24=0

a = 2; b = -20; c = -24;
Δ = b2-4ac
Δ = -202-4·2·(-24)
Δ = 592
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:

$\sqrt{\Delta}=\sqrt{592}=\sqrt{16*37}=\sqrt{16}*\sqrt{37}=4\sqrt{37}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-4\sqrt{37}}{2*2}=\frac{20-4\sqrt{37}}{4}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+4\sqrt{37}}{2*2}=\frac{20+4\sqrt{37}}{4}
$