2t^2=40

Solution for 2t^2=40 equation:

2t^2=40

We move all terms to the left:

2t^2-(40)=0

a = 2; b = 0; c = -40;
Δ = b2-4ac
Δ = 02-4·2·(-40)
Δ = 320
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{320}=\sqrt{64*5}=\sqrt{64}*\sqrt{5}=8\sqrt{5}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{5}}{2*2}=\frac{0-8\sqrt{5}}{4}
=-\frac{8\sqrt{5}}{4}
=-2\sqrt{5}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{5}}{2*2}=\frac{0+8\sqrt{5}}{4}
=\frac{8\sqrt{5}}{4}
=2\sqrt{5}
$