2t^2-12t+14=20

Solution for 2t^2-12t+14=20 equation:

2t^2-12t+14=20

We move all terms to the left:

2t^2-12t+14-(20)=0

We add all the numbers together, and all the variables

2t^2-12t-6=0

a = 2; b = -12; c = -6;
Δ = b2-4ac
Δ = -122-4·2·(-6)
Δ = 192
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:

$\sqrt{\Delta}=\sqrt{192}=\sqrt{64*3}=\sqrt{64}*\sqrt{3}=8\sqrt{3}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-8\sqrt{3}}{2*2}=\frac{12-8\sqrt{3}}{4}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+8\sqrt{3}}{2*2}=\frac{12+8\sqrt{3}}{4}
$