t^2+2t^2-2=0

Solution for t^2+2t^2-2=0 equation:

t^2+2t^2-2=0

We add all the numbers together, and all the variables

3t^2-2=0

a = 3; b = 0; c = -2;
Δ = b2-4ac
Δ = 02-4·3·(-2)
Δ = 24
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:

$\sqrt{\Delta}=\sqrt{24}=\sqrt{4*6}=\sqrt{4}*\sqrt{6}=2\sqrt{6}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{6}}{2*3}=\frac{0-2\sqrt{6}}{6}
=-\frac{2\sqrt{6}}{6}
=-\frac{\sqrt{6}}{3}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{6}}{2*3}=\frac{0+2\sqrt{6}}{6}
=\frac{2\sqrt{6}}{6}
=\frac{\sqrt{6}}{3}
$