-3t^2+56t-196=0

Solution for -3t^2+56t-196=0 equation:

-3t^2+56t-196=0

a = -3; b = 56; c = -196;
Δ = b2-4ac
Δ = 562-4·(-3)·(-196)
Δ = 784
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{784}=28$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(56)-28}{2*-3}=\frac{-84}{-6}
=+14
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(56)+28}{2*-3}=\frac{-28}{-6}
=4+2/3
$