2t^2=22

Solution for 2t^2=22 equation:

2t^2=22

We move all terms to the left:

2t^2-(22)=0

a = 2; b = 0; c = -22;
Δ = b2-4ac
Δ = 02-4·2·(-22)
Δ = 176
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:

$\sqrt{\Delta}=\sqrt{176}=\sqrt{16*11}=\sqrt{16}*\sqrt{11}=4\sqrt{11}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{11}}{2*2}=\frac{0-4\sqrt{11}}{4}
=-\frac{4\sqrt{11}}{4}
=-\sqrt{11}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{11}}{2*2}=\frac{0+4\sqrt{11}}{4}
=\frac{4\sqrt{11}}{4}
=\sqrt{11}
$