0.0625t^2-2t+10=0

Solution for 0.0625t^2-2t+10=0 equation:

0.0625t^2-2t+10=0

a = 0.0625; b = -2; c = +10;
Δ = b2-4ac
Δ = -22-4·0.0625·10
Δ = 1.5
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{1.5}}{2*0.0625}=\frac{2-\sqrt{1.5}}{0.125}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+\sqrt{1.5}}{2*0.0625}=\frac{2+\sqrt{1.5}}{0.125}
$