3t^2+20t-1200=0

Solution for 3t^2+20t-1200=0 equation:

3t^2+20t-1200=0

a = 3; b = 20; c = -1200;
Δ = b2-4ac
Δ = 202-4·3·(-1200)
Δ = 14800
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:

$\sqrt{\Delta}=\sqrt{14800}=\sqrt{400*37}=\sqrt{400}*\sqrt{37}=20\sqrt{37}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-20\sqrt{37}}{2*3}=\frac{-20-20\sqrt{37}}{6}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+20\sqrt{37}}{2*3}=\frac{-20+20\sqrt{37}}{6}
$