2t^2-16t+40=20

Solution for 2t^2-16t+40=20 equation:

2t^2-16t+40=20

We move all terms to the left:

2t^2-16t+40-(20)=0

We add all the numbers together, and all the variables

2t^2-16t+20=0

a = 2; b = -16; c = +20;
Δ = b2-4ac
Δ = -162-4·2·20
Δ = 96
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:

$\sqrt{\Delta}=\sqrt{96}=\sqrt{16*6}=\sqrt{16}*\sqrt{6}=4\sqrt{6}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-4\sqrt{6}}{2*2}=\frac{16-4\sqrt{6}}{4}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+4\sqrt{6}}{2*2}=\frac{16+4\sqrt{6}}{4}
$