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

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

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

a = -4; b = -2; c = +12;
Δ = b2-4ac
Δ = -22-4·(-4)·12
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-14}{2*-4}=\frac{-12}{-8}
=1+1/2
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+14}{2*-4}=\frac{16}{-8}
=-2
$