24+4t+t^2=180

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

24+4t+t^2=180

We move all terms to the left:

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

We add all the numbers together, and all the variables

t^2+4t-156=0

a = 1; b = 4; c = -156;
Δ = b2-4ac
Δ = 42-4·1·(-156)
Δ = 640
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:

$\sqrt{\Delta}=\sqrt{640}=\sqrt{64*10}=\sqrt{64}*\sqrt{10}=8\sqrt{10}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-8\sqrt{10}}{2*1}=\frac{-4-8\sqrt{10}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+8\sqrt{10}}{2*1}=\frac{-4+8\sqrt{10}}{2}
$