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

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

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

a = -12; b = 20; c = -3;
Δ = b2-4ac
Δ = 202-4·(-12)·(-3)
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-16}{2*-12}=\frac{-36}{-24}
=1+1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+16}{2*-12}=\frac{-4}{-24}
=1/6
$