z2+14z+40=

Solution for z2+14z+40= equation:

z2+14z+40=

We move all terms to the left:

z2+14z+40-()=0

We add all the numbers together, and all the variables

z^2+14z=0

a = 1; b = 14; c = 0;
Δ = b2-4ac
Δ = 142-4·1·0
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-14}{2*1}=\frac{-28}{2}
=-14
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+14}{2*1}=\frac{0}{2}
=0
$