2t2-16t-80=0

Solution for 2t2-16t-80=0 equation:

2t^2-16t-80=0

a = 2; b = -16; c = -80;
Δ = b2-4ac
Δ = -162-4·2·(-80)
Δ = 896
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:

$\sqrt{\Delta}=\sqrt{896}=\sqrt{64*14}=\sqrt{64}*\sqrt{14}=8\sqrt{14}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-8\sqrt{14}}{2*2}=\frac{16-8\sqrt{14}}{4}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+8\sqrt{14}}{2*2}=\frac{16+8\sqrt{14}}{4}
$