-.33t^2+103t+6279=0

Solution for -.33t^2+103t+6279=0 equation:

-.33t^2+103t+6279=0

We add all the numbers together, and all the variables

-0.33t^2+103t+6279=0

a = -0.33; b = 103; c = +6279;
Δ = b2-4ac
Δ = 1032-4·(-0.33)·6279
Δ = 18897.28
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(103)-\sqrt{18897.28}}{2*-0.33}=\frac{-103-\sqrt{18897.28}}{-0.66}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(103)+\sqrt{18897.28}}{2*-0.33}=\frac{-103+\sqrt{18897.28}}{-0.66}
$