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

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

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

We move all terms to the left:

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

We multiply parentheses

3y+2y-(8y-5y(y-10))-4=0


We calculate terms in parentheses: -(8y-5y(y-10)), so:

8y-5y(y-10)

We multiply parentheses

-5y^2+8y+50y

We add all the numbers together, and all the variables

-5y^2+58y

Back to the equation:

-(-5y^2+58y)


We add all the numbers together, and all the variables

-(-5y^2+58y)+5y-4=0

We get rid of parentheses

5y^2-58y+5y-4=0

We add all the numbers together, and all the variables

5y^2-53y-4=0

a = 5; b = -53; c = -4;
Δ = b2-4ac
Δ = -532-4·5·(-4)
Δ = 2889
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:

$\sqrt{\Delta}=\sqrt{2889}=\sqrt{9*321}=\sqrt{9}*\sqrt{321}=3\sqrt{321}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-53)-3\sqrt{321}}{2*5}=\frac{53-3\sqrt{321}}{10}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-53)+3\sqrt{321}}{2*5}=\frac{53+3\sqrt{321}}{10}
$