0=(-3t^2)+14t-14

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

0=(-3t^2)+14t-14

We move all terms to the left:

0-((-3t^2)+14t-14)=0

We add all the numbers together, and all the variables

-((-3t^2)+14t-14)=0


We calculate terms in parentheses: -((-3t^2)+14t-14), so:

(-3t^2)+14t-14

We get rid of parentheses

-3t^2+14t-14

Back to the equation:

-(-3t^2+14t-14)


We get rid of parentheses

3t^2-14t+14=0

a = 3; b = -14; c = +14;
Δ = b2-4ac
Δ = -142-4·3·14
Δ = 28
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:

$\sqrt{\Delta}=\sqrt{28}=\sqrt{4*7}=\sqrt{4}*\sqrt{7}=2\sqrt{7}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-2\sqrt{7}}{2*3}=\frac{14-2\sqrt{7}}{6}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+2\sqrt{7}}{2*3}=\frac{14+2\sqrt{7}}{6}
$