50=10t+t^2

Solution for 50=10t+t^2 equation:

50=10t+t^2

We move all terms to the left:

50-(10t+t^2)=0

We get rid of parentheses

-t^2-10t+50=0

We add all the numbers together, and all the variables

-1t^2-10t+50=0

a = -1; b = -10; c = +50;
Δ = b2-4ac
Δ = -102-4·(-1)·50
Δ = 300
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:

$\sqrt{\Delta}=\sqrt{300}=\sqrt{100*3}=\sqrt{100}*\sqrt{3}=10\sqrt{3}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-10\sqrt{3}}{2*-1}=\frac{10-10\sqrt{3}}{-2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+10\sqrt{3}}{2*-1}=\frac{10+10\sqrt{3}}{-2}
$