50=4t+t^2

Solution for 50=4t+t^2 equation:

50=4t+t^2

We move all terms to the left:

50-(4t+t^2)=0

We get rid of parentheses

-t^2-4t+50=0

We add all the numbers together, and all the variables

-1t^2-4t+50=0

a = -1; b = -4; c = +50;
Δ = b2-4ac
Δ = -42-4·(-1)·50
Δ = 216
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:

$\sqrt{\Delta}=\sqrt{216}=\sqrt{36*6}=\sqrt{36}*\sqrt{6}=6\sqrt{6}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-6\sqrt{6}}{2*-1}=\frac{4-6\sqrt{6}}{-2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+6\sqrt{6}}{2*-1}=\frac{4+6\sqrt{6}}{-2}
$