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

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

Solve for x (complex solution)

$$x=\frac{-\sqrt{15}i-5}{4}\approx -1.25-0.968245837i$$

$$x=1$$

$$x=\frac{-5+\sqrt{15}i}{4}\approx -1.25+0.968245837i$$

Solution Steps

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -5 and q divides the leading coefficient 2. List all candidates \frac{p}{q}.

$$±\frac{5}{2},±5,±\frac{1}{2},±1$$

Find one such root by trying out all the integer values, starting from the smallest by absolute value. If no integer roots are found, try out fractions.

$$x=1$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 2x^{3}+3x^{2}-5 by x-1 to get 2x^{2}+5x+5. Solve the equation where the result equals to 0.

$$2x^{2}+5x+5=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Substitute 2 for a, 5 for b, and 5 for c in the quadratic formula.

$$x=\frac{-5±\sqrt{5^{2}-4\times 2\times 5}}{2\times 2}$$

Do the calculations.

$$x=\frac{-5±\sqrt{-15}}{4}$$

Solve the equation 2x^{2}+5x+5=0 when ± is plus and when ± is minus.

$$x=\frac{-\sqrt{15}i-5}{4} x=\frac{-5+\sqrt{15}i}{4}$$

List all found solutions.

$$x=1 x=\frac{-\sqrt{15}i-5}{4} x=\frac{-5+\sqrt{15}i}{4}$$

Solve for x

$$x=1$$

Solution Steps

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -5 and q divides the leading coefficient 2. List all candidates \frac{p}{q}.

$$±\frac{5}{2},±5,±\frac{1}{2},±1$$

Find one such root by trying out all the integer values, starting from the smallest by absolute value. If no integer roots are found, try out fractions.

$$x=1$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 2x^{3}+3x^{2}-5 by x-1 to get 2x^{2}+5x+5. Solve the equation where the result equals to 0.

$$2x^{2}+5x+5=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Substitute 2 for a, 5 for b, and 5 for c in the quadratic formula.

$$x=\frac{-5±\sqrt{5^{2}-4\times 2\times 5}}{2\times 2}$$

Do the calculations.

$$x=\frac{-5±\sqrt{-15}}{4}$$

Since the square root of a negative number is not defined in the real field, there are no solutions.

$$x\in \emptyset$$

List all found solutions.

$$x=1$$