Answer:
[tex]x=\frac{1+\sqrt{31}}{2},\:x=\frac{1-\sqrt{31}}{2}[/tex]
Step-by-step explanation:
[tex]-> 0=2x^2-2x-15[/tex]
-zeros means the solution aka x intercept, meaning y=0
[tex]x_{1,\:2}=\frac{-\left(-2\right)\pm \sqrt{\left(-2\right)^2-4\cdot \:2\left(-15\right)}}{2\cdot \:2}[/tex]
=> according to quadratic formula
[tex]x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex]
where [tex]ax^2+bx+c=0[/tex]
=> [tex]x_{1,\:2}=\frac{-\left(-2\right)\pm \:2\sqrt{31}}{2\cdot \:2}[/tex]
simplify
[tex]x_1=\frac{-\left(-2\right)+2\sqrt{31}}{2\cdot \:2},\:x_2=\frac{-\left(-2\right)-2\sqrt{31}}{2\cdot \:2}[/tex]
[tex]x=\frac{1+\sqrt{31}}{2},\:x=\frac{1-\sqrt{31}}{2}[/tex]
Cannot simplify farther
