t2+2+8t-16=0

Solution for t2+2+8t-16=0 equation:

t2+2+8t-16=0

We add all the numbers together, and all the variables

t^2+8t-14=0

a = 1; b = 8; c = -14;
Δ = b2-4ac
Δ = 82-4·1·(-14)
Δ = 120
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:

$\sqrt{\Delta}=\sqrt{120}=\sqrt{4*30}=\sqrt{4}*\sqrt{30}=2\sqrt{30}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-2\sqrt{30}}{2*1}=\frac{-8-2\sqrt{30}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+2\sqrt{30}}{2*1}=\frac{-8+2\sqrt{30}}{2}
$