Question 255297
solve
25x^4-20x^2+3=0

substitute y for x^2

25y^2-20y+3=0
now we got a equation of form a*y^2 + b*y + c = 0
a=25, b=-20, c=3

using quadratic formula
{{{y = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
{{{y = (-(-20) +- sqrt( (-20)^2-4*25*3 ))/(2*25) }}}
{{{y = (20 +- sqrt( 400-300 ))/50 }}}
{{{y = (20 +- sqrt( 100 ))/50 }}}
{{{y = (20 +- 10)/50 }}}
{{{y1 = (20 + 10)/50 = 30/50 = 3/5}}}
{{{y2 = (20 - 10)/50 = 10/50 = 1/5}}}

y1=3/5,  y2=1/5

now going back to x
y=x^2
x=+- sqrt(y)

x1=+- sqrt(3/5)
x2=+- sqrt(1/5)