4t^2-36=20

Solution for 4t^2-36=20 equation:

4t^2-36=20

We move all terms to the left:

4t^2-36-(20)=0

We add all the numbers together, and all the variables

4t^2-56=0

a = 4; b = 0; c = -56;
Δ = b2-4ac
Δ = 02-4·4·(-56)
Δ = 896
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:

$\sqrt{\Delta}=\sqrt{896}=\sqrt{64*14}=\sqrt{64}*\sqrt{14}=8\sqrt{14}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{14}}{2*4}=\frac{0-8\sqrt{14}}{8}
=-\frac{8\sqrt{14}}{8}
=-\sqrt{14}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{14}}{2*4}=\frac{0+8\sqrt{14}}{8}
=\frac{8\sqrt{14}}{8}
=\sqrt{14}
$