t^2-14t+12=0

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

t^2-14t+12=0

a = 1; b = -14; c = +12;
Δ = b2-4ac
Δ = -142-4·1·12
Δ = 148
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:

$\sqrt{\Delta}=\sqrt{148}=\sqrt{4*37}=\sqrt{4}*\sqrt{37}=2\sqrt{37}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-2\sqrt{37}}{2*1}=\frac{14-2\sqrt{37}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+2\sqrt{37}}{2*1}=\frac{14+2\sqrt{37}}{2}
$