-5+5t-t^2=0

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

-5+5t-t^2=0

We add all the numbers together, and all the variables

-1t^2+5t-5=0

a = -1; b = 5; c = -5;
Δ = b2-4ac
Δ = 52-4·(-1)·(-5)
Δ = 5
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-\sqrt{5}}{2*-1}=\frac{-5-\sqrt{5}}{-2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+\sqrt{5}}{2*-1}=\frac{-5+\sqrt{5}}{-2}
$