3z2+-14z+16=0

Solution for 3z2+-14z+16=0 equation:

3z^2+-14z+16=0

We add all the numbers together, and all the variables

3z^2-14z=0

a = 3; b = -14; c = 0;
Δ = b2-4ac
Δ = -142-4·3·0
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*3}=\frac{0}{6}
=0
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+14}{2*3}=\frac{28}{6}
=4+2/3
$