z2+17z+16=0

Solution for z2+17z+16=0 equation:

z2+17z+16=0

We add all the numbers together, and all the variables

z^2+17z+16=0

a = 1; b = 17; c = +16;
Δ = b2-4ac
Δ = 172-4·1·16
Δ = 225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{225}=15$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(17)-15}{2*1}=\frac{-32}{2}
=-16
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(17)+15}{2*1}=\frac{-2}{2}
=-1
$