2z^2-18z+16=0

Solution for 2z^2-18z+16=0 equation:

2z^2-18z+16=0

a = 2; b = -18; c = +16;
Δ = b2-4ac
Δ = -182-4·2·16
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{-(-18)-14}{2*2}=\frac{4}{4}
=1
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+14}{2*2}=\frac{32}{4}
=8
$