3x^2-40x+33=-15

Solution for 3x^2-40x+33=-15 equation:

3x^2-40x+33=-15

We move all terms to the left:

3x^2-40x+33-(-15)=0

We add all the numbers together, and all the variables

3x^2-40x+48=0

a = 3; b = -40; c = +48;
Δ = b2-4ac
Δ = -402-4·3·48
Δ = 1024
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1024}=32$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-32}{2*3}=\frac{8}{6}
=1+1/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+32}{2*3}=\frac{72}{6}
=12
$