Solve $\color{orange}{a}x^2 - \color{blue}{b}x + \color{green}{c} = 0$

Generally, quadratic equation of the form $\color{orange}{a}x^2 - \color{blue}{b}x + \color{green}{c} = 0$ where $\color{orange}{a} \neq 0$ can be solve by evaluating $$\color{brown}{\Delta} = \color{blue}{b}^2 - 4\color{orange}{a}\color{green}{c}$$ Then, depending on the sign of $\color{brown}{\Delta}$ $$ \begin{array}{c|c|c} \color{brown}{\Delta} > 0 & \color{brown}{\Delta} = 0& \color{brown}{\Delta} < 0 \ \hline \textrm{two solutions} & \textrm{one solution} & \textrm{no real solutions}\ \hline x = \frac{-\color{blue}{b} \pm \sqrt{\color{brown}{\Delta}}}{2\color{orange}{a}} & x = \frac{-\color{blue}{b}}{2\color{orange}{a}}& - \end{array} $$ Then for your particular case we have $$\color{orange}{1}x^2 - \color{blue}{8}x + \color{green}{15} = 0$$ $$\color{brown}{\Delta} = (\color{blue}{-8})^2 - 4 \times \color{orange}{1} \times \color{green}{15}$$ $$\color{brown}{\Delta} = 4$$ Since $\color{brown}{\Delta} > 0$ we get that $$x = \frac{-(\color{blue}{-8}) \pm \sqrt{\color{brown}{4}}}{2 \times \color{orange}{1}}$$ $$x = \frac{8 \pm 2}{2}$$ $$x = 4 \pm 1$$ $$x = 5 \qquad \vee\ \qquad x = 3$$