3t^2-16t-400=0

Solution for 3t^2-16t-400=0 equation:

3t^2-16t-400=0

a = 3; b = -16; c = -400;
Δ = b2-4ac
Δ = -162-4·3·(-400)
Δ = 5056
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:

$\sqrt{\Delta}=\sqrt{5056}=\sqrt{64*79}=\sqrt{64}*\sqrt{79}=8\sqrt{79}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-8\sqrt{79}}{2*3}=\frac{16-8\sqrt{79}}{6}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+8\sqrt{79}}{2*3}=\frac{16+8\sqrt{79}}{6}
$