12+4z-z^2=0

Solution for 12+4z-z^2=0 equation:

12+4z-z^2=0

We add all the numbers together, and all the variables

-1z^2+4z+12=0

a = -1; b = 4; c = +12;
Δ = b2-4ac
Δ = 42-4·(-1)·12
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-8}{2*-1}=\frac{-12}{-2}
=+6
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+8}{2*-1}=\frac{4}{-2}
=-2
$