28=56t+t^2

Solution for 28=56t+t^2 equation:

28=56t+t^2

We move all terms to the left:

28-(56t+t^2)=0

We get rid of parentheses

-t^2-56t+28=0

We add all the numbers together, and all the variables

-1t^2-56t+28=0

a = -1; b = -56; c = +28;
Δ = b2-4ac
Δ = -562-4·(-1)·28
Δ = 3248
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:

$\sqrt{\Delta}=\sqrt{3248}=\sqrt{16*203}=\sqrt{16}*\sqrt{203}=4\sqrt{203}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-56)-4\sqrt{203}}{2*-1}=\frac{56-4\sqrt{203}}{-2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-56)+4\sqrt{203}}{2*-1}=\frac{56+4\sqrt{203}}{-2}
$