50+10t-5t^2=0

Solution for 50+10t-5t^2=0 equation:

50+10t-5t^2=0

a = -5; b = 10; c = +50;
Δ = b2-4ac
Δ = 102-4·(-5)·50
Δ = 1100
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:

$\sqrt{\Delta}=\sqrt{1100}=\sqrt{100*11}=\sqrt{100}*\sqrt{11}=10\sqrt{11}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-10\sqrt{11}}{2*-5}=\frac{-10-10\sqrt{11}}{-10}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+10\sqrt{11}}{2*-5}=\frac{-10+10\sqrt{11}}{-10}
$