12t^2+12t+14t-10t=12

Solution for 12t^2+12t+14t-10t=12 equation:

12t^2+12t+14t-10t=12

We move all terms to the left:

12t^2+12t+14t-10t-(12)=0

We add all the numbers together, and all the variables

12t^2+16t-12=0

a = 12; b = 16; c = -12;
Δ = b2-4ac
Δ = 162-4·12·(-12)
Δ = 832
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:

$\sqrt{\Delta}=\sqrt{832}=\sqrt{64*13}=\sqrt{64}*\sqrt{13}=8\sqrt{13}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-8\sqrt{13}}{2*12}=\frac{-16-8\sqrt{13}}{24}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+8\sqrt{13}}{2*12}=\frac{-16+8\sqrt{13}}{24}
$