32x^2-24+4=0

Solution for 32x^2-24+4=0 equation:

32x^2-24+4=0

We add all the numbers together, and all the variables

32x^2-20=0

a = 32; b = 0; c = -20;
Δ = b2-4ac
Δ = 02-4·32·(-20)
Δ = 2560
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:

$\sqrt{\Delta}=\sqrt{2560}=\sqrt{256*10}=\sqrt{256}*\sqrt{10}=16\sqrt{10}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-16\sqrt{10}}{2*32}=\frac{0-16\sqrt{10}}{64}
=-\frac{16\sqrt{10}}{64}
=-\frac{\sqrt{10}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+16\sqrt{10}}{2*32}=\frac{0+16\sqrt{10}}{64}
=\frac{16\sqrt{10}}{64}
=\frac{\sqrt{10}}{4}
$