33-5y=-8(3y+9)y=39/19

Solution for 33-5y=-8(3y+9)y=39/19 equation:

33-5y=-8(3y+9)y=39/19

We move all terms to the left:

33-5y-(-8(3y+9)y)=0


We calculate terms in parentheses: -(-8(3y+9)y), so:

-8(3y+9)y

We multiply parentheses

-24y^2-72y

Back to the equation:

-(-24y^2-72y)


We get rid of parentheses

24y^2+72y-5y+33=0

We add all the numbers together, and all the variables

24y^2+67y+33=0

a = 24; b = 67; c = +33;
Δ = b2-4ac
Δ = 672-4·24·33
Δ = 1321
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{-(67)-\sqrt{1321}}{2*24}=\frac{-67-\sqrt{1321}}{48}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(67)+\sqrt{1321}}{2*24}=\frac{-67+\sqrt{1321}}{48}
$