-t^2+20t-30=0

Solution for -t^2+20t-30=0 equation:

-t^2+20t-30=0

We add all the numbers together, and all the variables

-1t^2+20t-30=0

a = -1; b = 20; c = -30;
Δ = b2-4ac
Δ = 202-4·(-1)·(-30)
Δ = 280
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:

$\sqrt{\Delta}=\sqrt{280}=\sqrt{4*70}=\sqrt{4}*\sqrt{70}=2\sqrt{70}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-2\sqrt{70}}{2*-1}=\frac{-20-2\sqrt{70}}{-2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+2\sqrt{70}}{2*-1}=\frac{-20+2\sqrt{70}}{-2}
$