1z^2+10z+6=0

Solution for 1z^2+10z+6=0 equation:

1z^2+10z+6=0

We add all the numbers together, and all the variables

z^2+10z+6=0

a = 1; b = 10; c = +6;
Δ = b2-4ac
Δ = 102-4·1·6
Δ = 76
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:

$\sqrt{\Delta}=\sqrt{76}=\sqrt{4*19}=\sqrt{4}*\sqrt{19}=2\sqrt{19}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{19}}{2*1}=\frac{-10-2\sqrt{19}}{2}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{19}}{2*1}=\frac{-10+2\sqrt{19}}{2}
$