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

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 Click here to see ALL problems on real-numbers 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(60782)   (Show Source): You can put this solution on YOUR website!4x^2+5=6x^3 6x^3-4x^2-5 = 0 ----- I graphed it and found a Real solution at x = 1.223423... Cheers, Stan H. Answer by lwsshak3(7679)   (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(5390)   (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 (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.