64x^2=22

Solution for 64x^2=22 equation:

64x^2=22

We move all terms to the left:

64x^2-(22)=0

a = 64; b = 0; c = -22;
Δ = b2-4ac
Δ = 02-4·64·(-22)
Δ = 5632
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{5632}=\sqrt{256*22}=\sqrt{256}*\sqrt{22}=16\sqrt{22}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-16\sqrt{22}}{2*64}=\frac{0-16\sqrt{22}}{128}
=-\frac{16\sqrt{22}}{128}
=-\frac{\sqrt{22}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+16\sqrt{22}}{2*64}=\frac{0+16\sqrt{22}}{128}
=\frac{16\sqrt{22}}{128}
=\frac{\sqrt{22}}{8}
$