3t^2-14t+12=0

Solution for 3t^2-14t+12=0 equation:

3t^2-14t+12=0

a = 3; b = -14; c = +12;
Δ = b2-4ac
Δ = -142-4·3·12
Δ = 52
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:

$\sqrt{\Delta}=\sqrt{52}=\sqrt{4*13}=\sqrt{4}*\sqrt{13}=2\sqrt{13}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-2\sqrt{13}}{2*3}=\frac{14-2\sqrt{13}}{6}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+2\sqrt{13}}{2*3}=\frac{14+2\sqrt{13}}{6}
$