4(z^2+4z-3)=0

Solution for 4(z^2+4z-3)=0 equation:

4(z^2+4z-3)=0

We multiply parentheses

4z^2+16z-12=0

a = 4; b = 16; c = -12;
Δ = b2-4ac
Δ = 162-4·4·(-12)
Δ = 448
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:

$\sqrt{\Delta}=\sqrt{448}=\sqrt{64*7}=\sqrt{64}*\sqrt{7}=8\sqrt{7}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-8\sqrt{7}}{2*4}=\frac{-16-8\sqrt{7}}{8}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+8\sqrt{7}}{2*4}=\frac{-16+8\sqrt{7}}{8}
$