50+2t-0.0625t^2=0

Solution for 50+2t-0.0625t^2=0 equation:

50+2t-0.0625t^2=0

a = -0.0625; b = 2; c = +50;
Δ = b2-4ac
Δ = 22-4·(-0.0625)·50
Δ = 16.5
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{-(2)-\sqrt{16.5}}{2*-0.0625}=\frac{-2-\sqrt{16.5}}{-0.125}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+\sqrt{16.5}}{2*-0.0625}=\frac{-2+\sqrt{16.5}}{-0.125}
$