12x^2+8=40

Solution for 12x^2+8=40 equation:

12x^2+8=40

We move all terms to the left:

12x^2+8-(40)=0

We add all the numbers together, and all the variables

12x^2-32=0

a = 12; b = 0; c = -32;
Δ = b2-4ac
Δ = 02-4·12·(-32)
Δ = 1536
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:

$\sqrt{\Delta}=\sqrt{1536}=\sqrt{256*6}=\sqrt{256}*\sqrt{6}=16\sqrt{6}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-16\sqrt{6}}{2*12}=\frac{0-16\sqrt{6}}{24}
=-\frac{16\sqrt{6}}{24}
=-\frac{2\sqrt{6}}{3}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+16\sqrt{6}}{2*12}=\frac{0+16\sqrt{6}}{24}
=\frac{16\sqrt{6}}{24}
=\frac{2\sqrt{6}}{3}
$