3(2^2t-5)-4=10

Solution for 3(2^2t-5)-4=10 equation:

3(2^2t-5)-4=10

We move all terms to the left:

3(2^2t-5)-4-(10)=0

We add all the numbers together, and all the variables

3(2^2t-5)-14=0

We multiply parentheses

6t^2-15-14=0

We add all the numbers together, and all the variables

6t^2-29=0

a = 6; b = 0; c = -29;
Δ = b2-4ac
Δ = 02-4·6·(-29)
Δ = 696
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:

$\sqrt{\Delta}=\sqrt{696}=\sqrt{4*174}=\sqrt{4}*\sqrt{174}=2\sqrt{174}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{174}}{2*6}=\frac{0-2\sqrt{174}}{12}
=-\frac{2\sqrt{174}}{12}
=-\frac{\sqrt{174}}{6}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{174}}{2*6}=\frac{0+2\sqrt{174}}{12}
=\frac{2\sqrt{174}}{12}
=\frac{\sqrt{174}}{6}
$