Solve 0=w^3+3w^2-20

$$0 = w ^ { 3 } + 3 w ^ { 2 } - 20$$

Solve for w

$$w=2$$

Solution Steps

Swap sides so that all variable terms are on the left hand side.

$$w^{3}+3w^{2}-20=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -20 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±20,±10,±5,±4,±2,±1$$

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.

$$w=2$$

By Factor theorem, w-k is a factor of the polynomial for each root k. Divide w^{3}+3w^{2}-20 by w-2 to get w^{2}+5w+10. Solve the equation where the result equals to 0.

$$w^{2}+5w+10=0$$

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 1 for a, 5 for b, and 10 for c in the quadratic formula.

$$w=\frac{-5±\sqrt{5^{2}-4\times 1\times 10}}{2}$$

Do the calculations.

$$w=\frac{-5±\sqrt{-15}}{2}$$

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

$$w\in \emptyset$$

List all found solutions.

$$w=2$$