-5t^2+42t+100=0

Solution for -5t^2+42t+100=0 equation:

-5t^2+42t+100=0

a = -5; b = 42; c = +100;
Δ = b2-4ac
Δ = 422-4·(-5)·100
Δ = 3764
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:

$\sqrt{\Delta}=\sqrt{3764}=\sqrt{4*941}=\sqrt{4}*\sqrt{941}=2\sqrt{941}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(42)-2\sqrt{941}}{2*-5}=\frac{-42-2\sqrt{941}}{-10}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(42)+2\sqrt{941}}{2*-5}=\frac{-42+2\sqrt{941}}{-10}
$