0=-t2+4t+5

Solution for 0=-t2+4t+5 equation:

0=-t2+4t+5

We move all terms to the left:

0-(-t2+4t+5)=0

We add all the numbers together, and all the variables

-(-1t^2+4t+5)+0=0

We add all the numbers together, and all the variables

-(-1t^2+4t+5)=0

We get rid of parentheses

1t^2-4t-5=0

We add all the numbers together, and all the variables

t^2-4t-5=0

a = 1; b = -4; c = -5;
Δ = b2-4ac
Δ = -42-4·1·(-5)
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-6}{2*1}=\frac{-2}{2}
=-1
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+6}{2*1}=\frac{10}{2}
=5
$