2t^2+30t-125=0

Solution for 2t^2+30t-125=0 equation:

2t^2+30t-125=0

a = 2; b = 30; c = -125;
Δ = b2-4ac
Δ = 302-4·2·(-125)
Δ = 1900
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:

$\sqrt{\Delta}=\sqrt{1900}=\sqrt{100*19}=\sqrt{100}*\sqrt{19}=10\sqrt{19}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(30)-10\sqrt{19}}{2*2}=\frac{-30-10\sqrt{19}}{4}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(30)+10\sqrt{19}}{2*2}=\frac{-30+10\sqrt{19}}{4}
$