find all real solutions for the equation ( 15w - 18w^{2}+3w^{3}=0 ). separate multiple solutions with commas…

Explanation:

Step1: Factor out the common term

Factor out (3w) from the equation (15w - 18w^{2}+3w^{3}=0). We get (3w(w^{2}-6w + 5)=0). Since (3\neq0), we can focus on (w(w^{2}-6w + 5)=0).

Step2: Factor the quadratic expression

Factor (w^{2}-6w + 5). We know that (w^{2}-6w + 5=(w - 1)(w - 5)) (using the formula (ax^{2}+bx + c=a(x - x_1)(x - x_2)) where (x_{1,2}=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}) for (ax^{2}+bx + c = 0), here (a = 1), (b=-6), (c = 5), (x=\frac{6\pm\sqrt{36 - 20}}{2}=\frac{6\pm4}{2}), (x_1=1), (x_2 = 5)). So the equation becomes (w(w - 1)(w - 5)=0).

Step3: Apply the zero - product property

If (abc = 0), then (a = 0) or (b = 0) or (c = 0). For (w(w - 1)(w - 5)=0), we have: When (w=0), the equation holds. When (w - 1=0), then (w = 1). When (w - 5=0), then (w = 5).

Answer:

(0,1,5)