SOLUTION: iF I have 4x^2+5=6x^3 how would i solve for a real number solutions?

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Question 397940: iF I have 4x^2+5=6x^3 how would i solve for a real number solutions?
Found 3 solutions by stanbon, lwsshak3, richard1234:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
4x^2+5=6x^3
6x^3-4x^2-5 = 0
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I graphed it and found a Real solution at
x = 1.223423...
Cheers,
Stan H.

Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
iF I have 4x^2+5=6x^3 how would i solve for a real number solutions?
4x^2+5=6x^3
the only way I know how to get a real number solution is to do it graphically on a graphing calculator or a computer graphing program.
rearrange the equation and set it to zero
6x^3-4x^2-5=0
enter 6x^3-4x^2-5 into the graphing calculator and you will find that when y=0,
x=1.22342
The other two solutions must be imaginary because the curve crosses the x-axis only once as seen on the graph
Maybe a smarter tutor knows how to get an algebriac solution, but this one is beyond my pay grade.

Answer by richard1234(7193) About Me  (Show Source):
You can put this solution on YOUR website!
There is a cubic formula, but I wouldn't bother memorizing it...

You could use Newton's method to find the real zeros of the function. Newton's method defines a recursive sequence

x%5Bn%2B1%5D+=+x%5Bn%5D+-+f%28x%5Bn%5D%29%2Ff%5E%28%281%29%29+%28x%5Bn%5D%29 (the website doesn't let me use f'(x) so I used f^(1) (x))

and it can be iterated indefinitely to find real roots.