4z^2+4z-12=0

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

4z^2+4z-12=0

a = 4; b = 4; c = -12;
Δ = b2-4ac
Δ = 42-4·4·(-12)
Δ = 208
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:

$\sqrt{\Delta}=\sqrt{208}=\sqrt{16*13}=\sqrt{16}*\sqrt{13}=4\sqrt{13}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{13}}{2*4}=\frac{-4-4\sqrt{13}}{8}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{13}}{2*4}=\frac{-4+4\sqrt{13}}{8}
$