10t^2+20t-50=0

Solution for 10t^2+20t-50=0 equation:

10t^2+20t-50=0

a = 10; b = 20; c = -50;
Δ = b2-4ac
Δ = 202-4·10·(-50)
Δ = 2400
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:

$\sqrt{\Delta}=\sqrt{2400}=\sqrt{400*6}=\sqrt{400}*\sqrt{6}=20\sqrt{6}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-20\sqrt{6}}{2*10}=\frac{-20-20\sqrt{6}}{20}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+20\sqrt{6}}{2*10}=\frac{-20+20\sqrt{6}}{20}
$