80=2(t)^2-8t+16

Solution for 80=2(t)^2-8t+16 equation:

80=2(t)^2-8t+16

We move all terms to the left:

80-(2(t)^2-8t+16)=0

We get rid of parentheses

-2t^2+8t-16+80=0

We add all the numbers together, and all the variables

-2t^2+8t+64=0

a = -2; b = 8; c = +64;
Δ = b2-4ac
Δ = 82-4·(-2)·64
Δ = 576
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{576}=24$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-24}{2*-2}=\frac{-32}{-4}
=+8
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+24}{2*-2}=\frac{16}{-4}
=-4
$