7y+3y(y-4)=3(y+1)-2

Solution for 7y+3y(y-4)=3(y+1)-2 equation:

7y+3y(y-4)=3(y+1)-2

We move all terms to the left:

7y+3y(y-4)-(3(y+1)-2)=0

We multiply parentheses

3y^2+7y-12y-(3(y+1)-2)=0


We calculate terms in parentheses: -(3(y+1)-2), so:

3(y+1)-2

We multiply parentheses

3y+3-2

We add all the numbers together, and all the variables

3y+1

Back to the equation:

-(3y+1)


We add all the numbers together, and all the variables

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

3y^2-8y-1=0

a = 3; b = -8; c = -1;
Δ = b2-4ac
Δ = -82-4·3·(-1)
Δ = 76
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:

$\sqrt{\Delta}=\sqrt{76}=\sqrt{4*19}=\sqrt{4}*\sqrt{19}=2\sqrt{19}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-2\sqrt{19}}{2*3}=\frac{8-2\sqrt{19}}{6}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+2\sqrt{19}}{2*3}=\frac{8+2\sqrt{19}}{6}
$