3-9x^2+20x-12=0

Solution for 3-9x^2+20x-12=0 equation:

3-9x^2+20x-12=0

We add all the numbers together, and all the variables

-9x^2+20x-9=0

a = -9; b = 20; c = -9;
Δ = b2-4ac
Δ = 202-4·(-9)·(-9)
Δ = 76
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:

$\sqrt{\Delta}=\sqrt{76}=\sqrt{4*19}=\sqrt{4}*\sqrt{19}=2\sqrt{19}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-2\sqrt{19}}{2*-9}=\frac{-20-2\sqrt{19}}{-18}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+2\sqrt{19}}{2*-9}=\frac{-20+2\sqrt{19}}{-18}
$