(t^2+4)+24=180

Solution for (t^2+4)+24=180 equation:

(t^2+4)+24=180

We move all terms to the left:

(t^2+4)+24-(180)=0

We add all the numbers together, and all the variables

(t^2+4)-156=0

We get rid of parentheses

t^2+4-156=0

We add all the numbers together, and all the variables

t^2-152=0

a = 1; b = 0; c = -152;
Δ = b2-4ac
Δ = 02-4·1·(-152)
Δ = 608
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:

$\sqrt{\Delta}=\sqrt{608}=\sqrt{16*38}=\sqrt{16}*\sqrt{38}=4\sqrt{38}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{38}}{2*1}=\frac{0-4\sqrt{38}}{2}
=-\frac{4\sqrt{38}}{2}
=-2\sqrt{38}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{38}}{2*1}=\frac{0+4\sqrt{38}}{2}
=\frac{4\sqrt{38}}{2}
=2\sqrt{38}
$