## Intermediate Algebra: Connecting Concepts through Application

$\color{blue}{\left\{-4-\sqrt{46}, -4+\sqrt{46}\right\}}$
The solutions of the quadratic equation $ax^2+bx+c=0$ can be found using the quadratic formula: $$x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$$ Write the given equation in $ax^2+bx+c=0$ form to obtain: $$x^2+8x-30=0$$ The equation above has $a=1, b=8, \text{ and } c= -30$. Substitute these values into the quadratic formula to obtain: \begin{align*} x&=\frac{-8\pm\sqrt{8^2-4(1)(-30)}}{2(1)}\\\\ x&=\frac{-8\pm \sqrt{64+120}}{2}\\\\ x&=\frac{-8\pm \sqrt{184}}{2}\\\\ x&=\frac{-8\pm\sqrt{4(46)}}{2}\\\\ x&=\frac{-8\pm1\sqrt{46}}{2} \end{align*} Thus, $x_1=\dfrac{-8+2\sqrt{46}}{2}=-4+\sqrt{46}\\\\$ $x_2=\dfrac{-8-2\sqrt{46}}{2}=4+\sqrt{46}$ Therefore, the solution set is $\color{blue}{\left\{-4-\sqrt{46}, -4+\sqrt{46}\right\}}$.