25=-5t^2+20t+14

Solution for 25=-5t^2+20t+14 equation:

25=-5t^2+20t+14

We move all terms to the left:

25-(-5t^2+20t+14)=0

We get rid of parentheses

5t^2-20t-14+25=0

We add all the numbers together, and all the variables

5t^2-20t+11=0

a = 5; b = -20; c = +11;
Δ = b2-4ac
Δ = -202-4·5·11
Δ = 180
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:

$\sqrt{\Delta}=\sqrt{180}=\sqrt{36*5}=\sqrt{36}*\sqrt{5}=6\sqrt{5}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-6\sqrt{5}}{2*5}=\frac{20-6\sqrt{5}}{10}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+6\sqrt{5}}{2*5}=\frac{20+6\sqrt{5}}{10}
$