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

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

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

We move all terms to the left:

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

We multiply parentheses

3y^2+7y-24y-(3(y+2)-3)=0


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

3(y+2)-3

We multiply parentheses

3y+6-3

We add all the numbers together, and all the variables

3y+3

Back to the equation:

-(3y+3)


We add all the numbers together, and all the variables

3y^2-17y-(3y+3)=0

We get rid of parentheses

3y^2-17y-3y-3=0

We add all the numbers together, and all the variables

3y^2-20y-3=0

a = 3; b = -20; c = -3;
Δ = b2-4ac
Δ = -202-4·3·(-3)
Δ = 436
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:

$\sqrt{\Delta}=\sqrt{436}=\sqrt{4*109}=\sqrt{4}*\sqrt{109}=2\sqrt{109}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-2\sqrt{109}}{2*3}=\frac{20-2\sqrt{109}}{6}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+2\sqrt{109}}{2*3}=\frac{20+2\sqrt{109}}{6}
$