16z2+62=64z

Solution for 16z2+62=64z equation:

16z^2+62=64z

We move all terms to the left:

16z^2+62-(64z)=0

a = 16; b = -64; c = +62;
Δ = b2-4ac
Δ = -642-4·16·62
Δ = 128
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{128}=\sqrt{64*2}=\sqrt{64}*\sqrt{2}=8\sqrt{2}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-64)-8\sqrt{2}}{2*16}=\frac{64-8\sqrt{2}}{32}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-64)+8\sqrt{2}}{2*16}=\frac{64+8\sqrt{2}}{32}
$