0=-2t^2+12t+14

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

0=-2t^2+12t+14

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

2t^2-12t-14=0

a = 2; b = -12; c = -14;
Δ = b2-4ac
Δ = -122-4·2·(-14)
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-16}{2*2}=\frac{-4}{4}
=-1
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+16}{2*2}=\frac{28}{4}
=7
$