32z^2-72z+16=0

Solution for 32z^2-72z+16=0 equation:

32z^2-72z+16=0

a = 32; b = -72; c = +16;
Δ = b2-4ac
Δ = -722-4·32·16
Δ = 3136
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{3136}=56$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-72)-56}{2*32}=\frac{16}{64}
=1/4
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-72)+56}{2*32}=\frac{128}{64}
=2
$