2y+5y(y-8)=7y-2(y-10)

Solution for 2y+5y(y-8)=7y-2(y-10) equation:

2y+5y(y-8)=7y-2(y-10)

We move all terms to the left:

2y+5y(y-8)-(7y-2(y-10))=0

We multiply parentheses

5y^2+2y-40y-(7y-2(y-10))=0


We calculate terms in parentheses: -(7y-2(y-10)), so:

7y-2(y-10)

We multiply parentheses

7y-2y+20

We add all the numbers together, and all the variables

5y+20

Back to the equation:

-(5y+20)


We add all the numbers together, and all the variables

5y^2-38y-(5y+20)=0

We get rid of parentheses

5y^2-38y-5y-20=0

We add all the numbers together, and all the variables

5y^2-43y-20=0

a = 5; b = -43; c = -20;
Δ = b2-4ac
Δ = -432-4·5·(-20)
Δ = 2249
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-43)-\sqrt{2249}}{2*5}=\frac{43-\sqrt{2249}}{10}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-43)+\sqrt{2249}}{2*5}=\frac{43+\sqrt{2249}}{10}
$