64x^2-4=21

Solution for 64x^2-4=21 equation:

64x^2-4=21

We move all terms to the left:

64x^2-4-(21)=0

We add all the numbers together, and all the variables

64x^2-25=0

a = 64; b = 0; c = -25;
Δ = b2-4ac
Δ = 02-4·64·(-25)
Δ = 6400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{6400}=80$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-80}{2*64}=\frac{-80}{128}
=-5/8
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+80}{2*64}=\frac{80}{128}
=5/8
$