5y+3y(y-5)=5(y+1)-3

Solution for 5y+3y(y-5)=5(y+1)-3 equation:

5y+3y(y-5)=5(y+1)-3

We move all terms to the left:

5y+3y(y-5)-(5(y+1)-3)=0

We multiply parentheses

3y^2+5y-15y-(5(y+1)-3)=0


We calculate terms in parentheses: -(5(y+1)-3), so:

5(y+1)-3

We multiply parentheses

5y+5-3

We add all the numbers together, and all the variables

5y+2

Back to the equation:

-(5y+2)


We add all the numbers together, and all the variables

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

We get rid of parentheses

3y^2-10y-5y-2=0

We add all the numbers together, and all the variables

3y^2-15y-2=0

a = 3; b = -15; c = -2;
Δ = b2-4ac
Δ = -152-4·3·(-2)
Δ = 249
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{-(-15)-\sqrt{249}}{2*3}=\frac{15-\sqrt{249}}{6}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+\sqrt{249}}{2*3}=\frac{15+\sqrt{249}}{6}
$