Solve each quadratic in form equation.
1. 4y^4+9=13y^2
2. x-3x^1/22=0
3. (x-5)^2+2(x-5)-35=0
4. (x-2)^2-3x-2)+2=0
5. 2(x^2-5)^2-13(x^2-5)+20=0
6. 2x^2/3+5x^1/3-6=0
7. 6x^2/3-5x^1/3-6=0
8. x^-2+4x^-1=12
9. x^4-8x^2+7=0
10. x^4-2x^2-35=0
just make a substitution using, say, u to get a quadratic.
For example, in #6, let u = x^1/3 and you have
2u^2+5u-6=0
u = 1/4 (-5 ± √73)
Since u=x^1/3, x = u^3, so
x = 1/16 (-305 ± 37√73)
Kinda nasty, but straightforward.
How about #5? Let u=x^2-5, to get
2u^2 - 13u + 20 = 0
(u-4)(2u-5) = 0
so, u=4,5/2
That means
(x^2-5) = 4 or (x^2-5) = 5/2
x^2=9 or x^2 = 15/2
x = ±3 or ±√(15/2)