$0={x}^{3}+7{x}^{2}+7x-15$
Move the expression to the left-hand side and change its sign$0-{x}^{3}-7{x}^{2}-7x+15=0$
Removing $0$ doesn't change the value, so remove it from the expression$-{x}^{3}-7{x}^{2}-7x+15=0$
Write $-7{x}^{2}$ as a difference$-{x}^{3}+{x}^{2}-8{x}^{2}-7x+15=0$
Write $-7x$ as a difference$-{x}^{3}+{x}^{2}-8{x}^{2}+8x-15x+15=0$
Factor out $-{x}^{2}$ from the expression$-{x}^{2} \times \left( x-1 \right)-8{x}^{2}+8x-15x+15=0$
Factor out $-8x$ from the expression$-{x}^{2} \times \left( x-1 \right)-8x \times \left( x-1 \right)-15x+15=0$
Factor out $-15$ from the expression$-{x}^{2} \times \left( x-1 \right)-8x \times \left( x-1 \right)-15\left( x-1 \right)=0$
Factor out $-\left( x-1 \right)$ from the expression$-\left( x-1 \right) \times \left( {x}^{2}+8x+15 \right)=0$
Write $8x$ as a sum$-\left( x-1 \right) \times \left( {x}^{2}+5x+3x+15 \right)=0$
Factor out $x$ from the expression$-\left( x-1 \right) \times \left( x \times \left( x+5 \right)+3x+15 \right)=0$
Factor out $3$ from the expression$-\left( x-1 \right) \times \left( x \times \left( x+5 \right)+3\left( x+5 \right) \right)=0$
Factor out $x+5$ from the expression$-\left( x-1 \right) \times \left( x+5 \right) \times \left( x+3 \right)=0$
Change the signs on both sides of the equation$\left( x-1 \right) \times \left( x+5 \right) \times \left( x+3 \right)=0$
When the product of factors equals $0$, at least one factor is $0$$\begin{array} { l }x-1=0,\\x+5=0,\\x+3=0\end{array}$
Solve the equation for $x$$\begin{array} { l }x=1,\\x+5=0,\\x+3=0\end{array}$
Solve the equation for $x$$\begin{array} { l }x=1,\\x=-5,\\x+3=0\end{array}$
Solve the equation for $x$$\begin{array} { l }x=1,\\x=-5,\\x=-3\end{array}$
The equation has $3$ solutions$\begin{array} { l }x_1=-5,& x_2=-3,& x_3=1\end{array}$