Answer
Zeros: $\dfrac{5}{2},-\dfrac{3}{5}$
$x$-intercepts: $\dfrac{5}{2},-\dfrac{3}{5}$
Work Step by Step
To find the zeros of a function $f$, solve the equation $f(x)=0$
The zeros of the function are also the $x-$intercepts.
Let $G)x)=0$:
$$10x^2-19x-15=0$$
Comparing $10x^2-19x-15=0$ to $ax^2+bx+c=0$ to find $a,b \text{ and } c$
$$\therefore a = 10, b=-19 , c =-15$$
Evaluating the discriminant $b^2-4ac$
$$b^2-4ac = (-19)^2-4 \times 10 \times -15 = 961$$
The quadratic formula is given by:
$$x= \dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$$
$$x= \dfrac{19\pm \sqrt{961}}{2\times 10}$$
$$x=\dfrac{19\pm 31}{20}$$
Thus, the zeros, which are also the $x$-intercepts, are $\dfrac{5}{2}$ and $-\dfrac{3}{5}$.
