40=3x^2

Solution for 40=3x^2 equation:

40=3x^2

We move all terms to the left:

40-(3x^2)=0

a = -3; b = 0; c = +40;
Δ = b2-4ac
Δ = 02-4·(-3)·40
Δ = 480
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:

$\sqrt{\Delta}=\sqrt{480}=\sqrt{16*30}=\sqrt{16}*\sqrt{30}=4\sqrt{30}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{30}}{2*-3}=\frac{0-4\sqrt{30}}{-6}
=-\frac{4\sqrt{30}}{-6}
=-\frac{2\sqrt{30}}{-3}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{30}}{2*-3}=\frac{0+4\sqrt{30}}{-6}
=\frac{4\sqrt{30}}{-6}
=\frac{2\sqrt{30}}{-3}
$