Solve 0 = 0.7x + 6.804e-10 * x^4 - 1210.23 for all real x.

Step-by-step solution

1. 1

   Write the quartic in standard form. Reordering by degree,

   $6.804\times10^{-10}\,x^{4}+0.7x-1210.23=0.$

   There are no $x^{3}$ or $x^{2}$ terms, but that does not make the equation solvable by substitution — the odd $x^{1}$ term rules out a biquadratic reduction.

2. 2

   Resist the temptation to drop the quartic term. Because $6.804\times10^{-10}$ looks negligible it is tempting to solve $0.7x=1210.23$ and quote $x\approx 1728.9$. That is wrong: at $x=1728.9$ the quartic term equals

   $6.804\times10^{-10}\times(1728.9)^{4}\approx 6079,$

   which is five times the constant $1210.23$, not a small correction. A small coefficient is not a small term once $x$ is large — the crossover here is around $x\approx 1000$, well below the naive estimate.

3. 3

   Bracket the roots by sampling. Evaluating $f(x)=6.804\times10^{-10}x^{4}+0.7x-1210.23$:

   $f(0)=-1210.23,\quad f(500)=-817.71,\quad f(1000)=+170.17,\quad f(1500)=+3284.30.$

   So a root lies in $(500,1000)$. Sweeping across $[-5000,5000]$ finds exactly two sign changes: one near $947$ and one near $-1332$ — the quartic has two real roots and two complex ones.

4. 4

   Bisect each bracket. Halving repeatedly until $|f|<10^{-6}$:

   $x_{+}\approx 947.035581,\qquad x_{-}\approx-1332.142808.$

   Newton's method from $x_{0}=1000$ converges to the same positive value in three iterations, using $f'(x)=2.7216\times10^{-9}x^{3}+0.7$.

5. 5

   Substitute both back. At $x=947.035581$: the quartic term is $547.31$ and the linear term is $662.92$, summing with $-1210.23$ to $\approx-3\times10^{-7}$ ✓. At $x=-1332.142808$: the quartic term is $2142.73$ and the linear term is $-932.50$, again cancelling the constant ✓. Note how both terms are comparable in size at the true roots — the opposite of the assumption that produced $1728.9$.