3x^2-19x^2+33x-9=0

Solution for 3x^2-19x^2+33x-9=0 equation:

3x^2-19x^2+33x-9=0

We add all the numbers together, and all the variables

-16x^2+33x-9=0

a = -16; b = 33; c = -9;
Δ = b2-4ac
Δ = 332-4·(-16)·(-9)
Δ = 513
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{513}=\sqrt{9*57}=\sqrt{9}*\sqrt{57}=3\sqrt{57}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(33)-3\sqrt{57}}{2*-16}=\frac{-33-3\sqrt{57}}{-32}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(33)+3\sqrt{57}}{2*-16}=\frac{-33+3\sqrt{57}}{-32}
$