24=4t+t^2

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

24=4t+t^2

We move all terms to the left:

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

We get rid of parentheses

-t^2-4t+24=0

We add all the numbers together, and all the variables

-1t^2-4t+24=0

a = -1; b = -4; c = +24;
Δ = b2-4ac
Δ = -42-4·(-1)·24
Δ = 112
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:

$\sqrt{\Delta}=\sqrt{112}=\sqrt{16*7}=\sqrt{16}*\sqrt{7}=4\sqrt{7}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4\sqrt{7}}{2*-1}=\frac{4-4\sqrt{7}}{-2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4\sqrt{7}}{2*-1}=\frac{4+4\sqrt{7}}{-2}
$