Quadratic polynomial can be factored using the transformation $ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)$, where $x_{1}$ and $x_{2}$ are the solutions of the quadratic equation $ax^{2}+bx+c=0$.
$$z^{2}+7z-3=0$$
All equations of the form $ax^{2}+bx+c=0$ can be solved using the quadratic formula: $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$. The quadratic formula gives two solutions, one when $±$ is addition and one when it is subtraction.
$$z=\frac{-7±\sqrt{7^{2}-4\left(-3\right)}}{2}$$
Square $7$.
$$z=\frac{-7±\sqrt{49-4\left(-3\right)}}{2}$$
Multiply $-4$ times $-3$.
$$z=\frac{-7±\sqrt{49+12}}{2}$$
Add $49$ to $12$.
$$z=\frac{-7±\sqrt{61}}{2}$$
Now solve the equation $z=\frac{-7±\sqrt{61}}{2}$ when $±$ is plus. Add $-7$ to $\sqrt{61}$.
$$z=\frac{\sqrt{61}-7}{2}$$
Now solve the equation $z=\frac{-7±\sqrt{61}}{2}$ when $±$ is minus. Subtract $\sqrt{61}$ from $-7$.
$$z=\frac{-\sqrt{61}-7}{2}$$
Factor the original expression using $ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)$. Substitute $\frac{-7+\sqrt{61}}{2}$ for $x_{1}$ and $\frac{-7-\sqrt{61}}{2}$ for $x_{2}$.
