0=-5t^2+20t+14

Solution for 0=-5t^2+20t+14 equation:

0=-5t^2+20t+14

We move all terms to the left:

0-(-5t^2+20t+14)=0

We add all the numbers together, and all the variables

-(-5t^2+20t+14)=0

We get rid of parentheses

5t^2-20t-14=0

a = 5; b = -20; c = -14;
Δ = b2-4ac
Δ = -202-4·5·(-14)
Δ = 680
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:

$\sqrt{\Delta}=\sqrt{680}=\sqrt{4*170}=\sqrt{4}*\sqrt{170}=2\sqrt{170}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-2\sqrt{170}}{2*5}=\frac{20-2\sqrt{170}}{10}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+2\sqrt{170}}{2*5}=\frac{20+2\sqrt{170}}{10}
$