2t^2-14t+3=0

Solution for 2t^2-14t+3=0 equation:

2t^2-14t+3=0

a = 2; b = -14; c = +3;
Δ = b2-4ac
Δ = -142-4·2·3
Δ = 172
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:

$\sqrt{\Delta}=\sqrt{172}=\sqrt{4*43}=\sqrt{4}*\sqrt{43}=2\sqrt{43}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-2\sqrt{43}}{2*2}=\frac{14-2\sqrt{43}}{4}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+2\sqrt{43}}{2*2}=\frac{14+2\sqrt{43}}{4}
$