0.35t^2-50t+100=0

Solution for 0.35t^2-50t+100=0 equation:

0.35t^2-50t+100=0

a = 0.35; b = -50; c = +100;
Δ = b2-4ac
Δ = -502-4·0.35·100
Δ = 2360
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:

$\sqrt{\Delta}=\sqrt{2360}=\sqrt{4*590}=\sqrt{4}*\sqrt{590}=2\sqrt{590}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-50)-2\sqrt{590}}{2*0.35}=\frac{50-2\sqrt{590}}{0.7}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-50)+2\sqrt{590}}{2*0.35}=\frac{50+2\sqrt{590}}{0.7}
$