2t^2+25t-200=0

Solution for 2t^2+25t-200=0 equation:

2t^2+25t-200=0

a = 2; b = 25; c = -200;
Δ = b2-4ac
Δ = 252-4·2·(-200)
Δ = 2225
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:

$\sqrt{\Delta}=\sqrt{2225}=\sqrt{25*89}=\sqrt{25}*\sqrt{89}=5\sqrt{89}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(25)-5\sqrt{89}}{2*2}=\frac{-25-5\sqrt{89}}{4}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(25)+5\sqrt{89}}{2*2}=\frac{-25+5\sqrt{89}}{4}
$