0.1t^2-10t-62.5=0

Solution for 0.1t^2-10t-62.5=0 equation:

0.1t^2-10t-62.5=0

a = 0.1; b = -10; c = -62.5;
Δ = b2-4ac
Δ = -102-4·0.1·(-62.5)
Δ = 125
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:

$\sqrt{\Delta}=\sqrt{125}=\sqrt{25*5}=\sqrt{25}*\sqrt{5}=5\sqrt{5}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-5\sqrt{5}}{2*0.1}=\frac{10-5\sqrt{5}}{0.2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+5\sqrt{5}}{2*0.1}=\frac{10+5\sqrt{5}}{0.2}
$