32+4t-t^2=0

Solution for 32+4t-t^2=0 equation:

32+4t-t^2=0

We add all the numbers together, and all the variables

-1t^2+4t+32=0

a = -1; b = 4; c = +32;
Δ = b2-4ac
Δ = 42-4·(-1)·32
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-12}{2*-1}=\frac{-16}{-2}
=+8
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+12}{2*-1}=\frac{8}{-2}
=-4
$