33x^2-40x+12=0

Solution for 33x^2-40x+12=0 equation:

33x^2-40x+12=0

a = 33; b = -40; c = +12;
Δ = b2-4ac
Δ = -402-4·33·12
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-4}{2*33}=\frac{36}{66}
=6/11
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+4}{2*33}=\frac{44}{66}
=2/3
$