5y+3y(y-2)=5(y+1)-3

Solution for 5y+3y(y-2)=5(y+1)-3 equation:

5y+3y(y-2)=5(y+1)-3

We move all terms to the left:

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

We multiply parentheses

3y^2+5y-6y-(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^2-1y-(5y+2)=0

We get rid of parentheses

3y^2-1y-5y-2=0

We add all the numbers together, and all the variables

3y^2-6y-2=0

a = 3; b = -6; c = -2;
Δ = b2-4ac
Δ = -62-4·3·(-2)
Δ = 60
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:

$\sqrt{\Delta}=\sqrt{60}=\sqrt{4*15}=\sqrt{4}*\sqrt{15}=2\sqrt{15}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-2\sqrt{15}}{2*3}=\frac{6-2\sqrt{15}}{6}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+2\sqrt{15}}{2*3}=\frac{6+2\sqrt{15}}{6}
$