5(y-3)-8=-7y(2y+5)-y

Solution for 5(y-3)-8=-7y(2y+5)-y equation:

5(y-3)-8=-7y(2y+5)-y

We move all terms to the left:

5(y-3)-8-(-7y(2y+5)-y)=0

We multiply parentheses

5y-(-7y(2y+5)-y)-15-8=0


We calculate terms in parentheses: -(-7y(2y+5)-y), so:

-7y(2y+5)-y

We add all the numbers together, and all the variables

-1y-7y(2y+5)

We multiply parentheses

-14y^2-1y-35y

We add all the numbers together, and all the variables

-14y^2-36y

Back to the equation:

-(-14y^2-36y)


We add all the numbers together, and all the variables

-(-14y^2-36y)+5y-23=0

We get rid of parentheses

14y^2+36y+5y-23=0

We add all the numbers together, and all the variables

14y^2+41y-23=0

a = 14; b = 41; c = -23;
Δ = b2-4ac
Δ = 412-4·14·(-23)
Δ = 2969
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{-(41)-\sqrt{2969}}{2*14}=\frac{-41-\sqrt{2969}}{28}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(41)+\sqrt{2969}}{2*14}=\frac{-41+\sqrt{2969}}{28}
$