z^2+16z+61=0

Solution for z^2+16z+61=0 equation:

z^2+16z+61=0

a = 1; b = 16; c = +61;
Δ = b2-4ac
Δ = 162-4·1·61
Δ = 12
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:

$\sqrt{\Delta}=\sqrt{12}=\sqrt{4*3}=\sqrt{4}*\sqrt{3}=2\sqrt{3}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-2\sqrt{3}}{2*1}=\frac{-16-2\sqrt{3}}{2}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+2\sqrt{3}}{2*1}=\frac{-16+2\sqrt{3}}{2}
$