((i-2)^3-5*i)/(6*i)+i+2
a^4 + 12a^2 + 35
$X^{2}$ - 22X -64 = 0
71a^2+44a+4=0
A sequence is such that the sum of its any number of terms, beginning from the first, is four times as large as the square of the ...
2+2
a^4 + 12x^2 + 35
a=1/b=✓3+✓5 if
a^5b^2-8a^2b^5
2+2=
x+19=2(x+6)
