-t^2+8t+33=0

Solution for -t^2+8t+33=0 equation:

-t^2+8t+33=0

We add all the numbers together, and all the variables

-1t^2+8t+33=0

a = -1; b = 8; c = +33;
Δ = b2-4ac
Δ = 82-4·(-1)·33
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-14}{2*-1}=\frac{-22}{-2}
=+11
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+14}{2*-1}=\frac{6}{-2}
=-3
$