Respuesta :
[tex]y=2x^2-7x-18[/tex]
In this problem, we want to find the zeros for the given function above.
Unfortunately it is not factorable, so we will need to use the quadratic formula:
[tex]x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]Recall that the zeros of a quadratic function are the same as the solutions or x-intercepts.
From our equation, we can get the values of a, b, and c to use in the quadratic formula:
[tex]\begin{gathered} y=2x^2-7x-18 \\ \\ a=2,b=-7,c=-18 \end{gathered}[/tex]When we substitute those values, we get:
[tex]x=\frac{-(-7)\pm\sqrt{(-7)^2-4(2)(-18)}}{2(2)}[/tex]Simplify the numerator and denominator as much as possible:
[tex]x=\frac{7\pm\sqrt{49+144}}{4}=\frac{7\pm\sqrt{193}}{4}[/tex]Split this into two equations, the addition and subtraction equations:
[tex]x=\frac{7+\sqrt{193}}{4},\text{ and }x=\frac{7-\sqrt{193}}{4}[/tex]We can use a calculator to get the final values:
[tex]\begin{gathered} x=\frac{7+\sqrt{193}}{4}\approx5.22 \\ \\ x=\frac{7-\sqrt{193}}{4}\approx-1.72 \end{gathered}[/tex]Our zeros are:
[tex]\boxed{5.22\text{ and }-1.72}[/tex]