Question of finding the value of x

Question

Find the value of $x$ if

$$x^3-2x^2+3x+5=0$$

Note it's values, there are 3 possible solutions as it is a degree 3 polynomial. — Physical Mathematics, Apr 18 '20 at 17:46
Does this answer your question? Is there anything like “cubic formula”? — Physical Mathematics, Apr 18 '20 at 17:47
https://www.wolframalpha.com/input/?i=x%5E3-2x%5E2%2B3x%2B5%3D0&t=crmtb01 — saulspatz, Apr 18 '20 at 17:50
It is the same equation and it is inside the another question — KHAN SAHAB, Apr 18 '20 at 17:59
"there are 3 possible solutions as it is a degree 3 polynomial" There could be as many as 3 possible solutions but it doesn't have to have as many as three. $x^3 + x=x(x^2+1)= 0$ for instance only has one solutions. — fleablood, Apr 18 '20 at 17:59
I didn't understand to find out one value of x so that in could be in quadratic form and I can easily find out the other two value... — KHAN SAHAB, Apr 18 '20 at 18:03

score 1 · Accepted Answer · answered Apr 18 '20 at 18:18

If $ax^3+bx^2+cx+d=0$, then the real solution of the equation is:

$x_1=$

$-\frac{b}{3a}$

$+\sqrt[3]{\left ( \frac{-b^3}{27a^3}+\frac{bc}{6a^2}-\frac{d}{2a} \right )+\sqrt{\left ( \frac{-b^3}{27a^3}+\frac{bc}{6a^2}-\frac{d}{2a} \right )^2+\left ( \frac{c}{3a}-\frac{b^2}{9a^2} \right )^3}}$

$+\sqrt[3]{\left ( \frac{-b^3}{27a^3}+\frac{bc}{6a^2}-\frac{d}{2a} \right )-\sqrt{\left ( \frac{-b^3}{27a^3}+\frac{bc}{6a^2}-\frac{d}{2a} \right )^2+\left ( \frac{c}{3a}-\frac{b^2}{9a^2} \right )^3}}$

In your case, put $a=1,b=-2,c=3,d=5$, you will get the real value of $x$:

The other two roots are complex, the can be obtained by factoring the left hand side of your equation, resulting a quadratic factor, which can be solved using the formula:

If $Ax^2+Bx+C=0$, and $A \ne 0$, then:

$$x_{2,3}=\frac{-B \pm \sqrt{B^2-4AC}}{2A}$$

Question of finding the value of x

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