z2+26z+35=0

Solution for z2+26z+35=0 equation:

z2+26z+35=0

We add all the numbers together, and all the variables

z^2+26z+35=0

a = 1; b = 26; c = +35;
Δ = b2-4ac
Δ = 262-4·1·35
Δ = 536
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:

$\sqrt{\Delta}=\sqrt{536}=\sqrt{4*134}=\sqrt{4}*\sqrt{134}=2\sqrt{134}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(26)-2\sqrt{134}}{2*1}=\frac{-26-2\sqrt{134}}{2}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(26)+2\sqrt{134}}{2*1}=\frac{-26+2\sqrt{134}}{2}
$