50=5t+0.05t^2

Solution for 50=5t+0.05t^2 equation:

50=5t+0.05t^2

We move all terms to the left:

50-(5t+0.05t^2)=0

We get rid of parentheses

-0.05t^2-5t+50=0

a = -0.05; b = -5; c = +50;
Δ = b2-4ac
Δ = -52-4·(-0.05)·50
Δ = 35
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-\sqrt{35}}{2*-0.05}=\frac{5-\sqrt{35}}{-0.1}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+\sqrt{35}}{2*-0.05}=\frac{5+\sqrt{35}}{-0.1}
$