3x^2-32+12=0

Solution for 3x^2-32+12=0 equation:

3x^2-32+12=0

We add all the numbers together, and all the variables

3x^2-20=0

a = 3; b = 0; c = -20;
Δ = b2-4ac
Δ = 02-4·3·(-20)
Δ = 240
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:

$\sqrt{\Delta}=\sqrt{240}=\sqrt{16*15}=\sqrt{16}*\sqrt{15}=4\sqrt{15}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{15}}{2*3}=\frac{0-4\sqrt{15}}{6}
=-\frac{4\sqrt{15}}{6}
=-\frac{2\sqrt{15}}{3}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{15}}{2*3}=\frac{0+4\sqrt{15}}{6}
=\frac{4\sqrt{15}}{6}
=\frac{2\sqrt{15}}{3}
$