(2t^2-25t-250)=0

Solution for (2t^2-25t-250)=0 equation:

(2t^2-25t-250)=0

We get rid of parentheses

2t^2-25t-250=0

a = 2; b = -25; c = -250;
Δ = b2-4ac
Δ = -252-4·2·(-250)
Δ = 2625
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:

$\sqrt{\Delta}=\sqrt{2625}=\sqrt{25*105}=\sqrt{25}*\sqrt{105}=5\sqrt{105}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-25)-5\sqrt{105}}{2*2}=\frac{25-5\sqrt{105}}{4}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-25)+5\sqrt{105}}{2*2}=\frac{25+5\sqrt{105}}{4}
$