4z^2+16z+11=0

Solution for 4z^2+16z+11=0 equation:

4z^2+16z+11=0

a = 4; b = 16; c = +11;
Δ = b2-4ac
Δ = 162-4·4·11
Δ = 80
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:

$\sqrt{\Delta}=\sqrt{80}=\sqrt{16*5}=\sqrt{16}*\sqrt{5}=4\sqrt{5}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-4\sqrt{5}}{2*4}=\frac{-16-4\sqrt{5}}{8}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+4\sqrt{5}}{2*4}=\frac{-16+4\sqrt{5}}{8}
$