2t^2-11=0

Solution for 2t^2-11=0 equation:

2t^2-11=0

a = 2; b = 0; c = -11;
Δ = b2-4ac
Δ = 02-4·2·(-11)
Δ = 88
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:

$\sqrt{\Delta}=\sqrt{88}=\sqrt{4*22}=\sqrt{4}*\sqrt{22}=2\sqrt{22}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{22}}{2*2}=\frac{0-2\sqrt{22}}{4}
=-\frac{2\sqrt{22}}{4}
=-\frac{\sqrt{22}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{22}}{2*2}=\frac{0+2\sqrt{22}}{4}
=\frac{2\sqrt{22}}{4}
=\frac{\sqrt{22}}{2}
$