t^2+80t-1344=0

Solution for t^2+80t-1344=0 equation:

t^2+80t-1344=0

a = 1; b = 80; c = -1344;
Δ = b2-4ac
Δ = 802-4·1·(-1344)
Δ = 11776
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:

$\sqrt{\Delta}=\sqrt{11776}=\sqrt{256*46}=\sqrt{256}*\sqrt{46}=16\sqrt{46}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(80)-16\sqrt{46}}{2*1}=\frac{-80-16\sqrt{46}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(80)+16\sqrt{46}}{2*1}=\frac{-80+16\sqrt{46}}{2}
$