z^2+10z+1=0

Solution for z^2+10z+1=0 equation:

z^2+10z+1=0

a = 1; b = 10; c = +1;
Δ = b2-4ac
Δ = 102-4·1·1
Δ = 96
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:

$\sqrt{\Delta}=\sqrt{96}=\sqrt{16*6}=\sqrt{16}*\sqrt{6}=4\sqrt{6}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-4\sqrt{6}}{2*1}=\frac{-10-4\sqrt{6}}{2}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+4\sqrt{6}}{2*1}=\frac{-10+4\sqrt{6}}{2}
$