4z^2+2z-16=0

Solution for 4z^2+2z-16=0 equation:

4z^2+2z-16=0

a = 4; b = 2; c = -16;
Δ = b2-4ac
Δ = 22-4·4·(-16)
Δ = 260
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{260}=\sqrt{4*65}=\sqrt{4}*\sqrt{65}=2\sqrt{65}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{65}}{2*4}=\frac{-2-2\sqrt{65}}{8}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{65}}{2*4}=\frac{-2+2\sqrt{65}}{8}
$