-50t^2+40+100=0

Solution for -50t^2+40+100=0 equation:

-50t^2+40+100=0

We add all the numbers together, and all the variables

-50t^2+140=0

a = -50; b = 0; c = +140;
Δ = b2-4ac
Δ = 02-4·(-50)·140
Δ = 28000
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:

$\sqrt{\Delta}=\sqrt{28000}=\sqrt{400*70}=\sqrt{400}*\sqrt{70}=20\sqrt{70}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-20\sqrt{70}}{2*-50}=\frac{0-20\sqrt{70}}{-100}
=-\frac{20\sqrt{70}}{-100}
=-\frac{\sqrt{70}}{-5}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+20\sqrt{70}}{2*-50}=\frac{0+20\sqrt{70}}{-100}
=\frac{20\sqrt{70}}{-100}
=\frac{\sqrt{70}}{-5}
$