2t^2-4=1

Solution for 2t^2-4=1 equation:

2t^2-4=1

We move all terms to the left:

2t^2-4-(1)=0

We add all the numbers together, and all the variables

2t^2-5=0

a = 2; b = 0; c = -5;
Δ = b2-4ac
Δ = 02-4·2·(-5)
Δ = 40
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:

$\sqrt{\Delta}=\sqrt{40}=\sqrt{4*10}=\sqrt{4}*\sqrt{10}=2\sqrt{10}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{10}}{2*2}=\frac{0-2\sqrt{10}}{4}
=-\frac{2\sqrt{10}}{4}
=-\frac{\sqrt{10}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{10}}{2*2}=\frac{0+2\sqrt{10}}{4}
=\frac{2\sqrt{10}}{4}
=\frac{\sqrt{10}}{2}
$