t^2-8t+8=0

Solution for t^2-8t+8=0 equation:

t^2-8t+8=0

a = 1; b = -8; c = +8;
Δ = b2-4ac
Δ = -82-4·1·8
Δ = 32
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:

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