-10t^2+50t+-35=0

Solution for -10t^2+50t+-35=0 equation:

-10t^2+50t+-35=0

We add all the numbers together, and all the variables

-10t^2+50t=0

a = -10; b = 50; c = 0;
Δ = b2-4ac
Δ = 502-4·(-10)·0
Δ = 2500
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2500}=50$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(50)-50}{2*-10}=\frac{-100}{-20}
=+5
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(50)+50}{2*-10}=\frac{0}{-20}
=0
$