2t^2+16t-6=0

Solution for 2t^2+16t-6=0 equation:

2t^2+16t-6=0

a = 2; b = 16; c = -6;
Δ = b2-4ac
Δ = 162-4·2·(-6)
Δ = 304
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:

$\sqrt{\Delta}=\sqrt{304}=\sqrt{16*19}=\sqrt{16}*\sqrt{19}=4\sqrt{19}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-4\sqrt{19}}{2*2}=\frac{-16-4\sqrt{19}}{4}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+4\sqrt{19}}{2*2}=\frac{-16+4\sqrt{19}}{4}
$