20=t^2-30t+45

Solution for 20=t^2-30t+45 equation:

20=t^2-30t+45

We move all terms to the left:

20-(t^2-30t+45)=0

We get rid of parentheses

-t^2+30t-45+20=0

We add all the numbers together, and all the variables

-1t^2+30t-25=0

a = -1; b = 30; c = -25;
Δ = b2-4ac
Δ = 302-4·(-1)·(-25)
Δ = 800
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:

$\sqrt{\Delta}=\sqrt{800}=\sqrt{400*2}=\sqrt{400}*\sqrt{2}=20\sqrt{2}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(30)-20\sqrt{2}}{2*-1}=\frac{-30-20\sqrt{2}}{-2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(30)+20\sqrt{2}}{2*-1}=\frac{-30+20\sqrt{2}}{-2}
$