#### Answer

$x=\left\{ -2,5 \right\}$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
Express the given equation, $
x^2-3x-10=0
,$ in factored form. Then, use the Zero-Factor Property by equating each factor to zero. Finally, solve each equation.
$\bf{\text{Solution Details:}}$
The factored form of the equation above is
\begin{array}{l}\require{cancel}
(x+2)(x-5)=0
.\end{array}
Equating each factor to zero (Zero-Factor Property), then
\begin{array}{l}\require{cancel}
x+2=0
\text{ OR }
x-5=0
.\end{array}
Using the properties of equality to solve each of the equation above results to
\begin{array}{l}\require{cancel}
x+2=0
\\\\
x=-2
\\\\\text{ OR }\\\\
x-5=0
\\\\
x=5
.\end{array}
Hence, the solutions are $
x=\left\{ -2,5 \right\}
.$
$\bf{\text{Supplementary Solution:}}$
To factor the expression, $
x^2-3x-10
,$ find two numbers, $m_1$ and $m_2,$ whose product is $c$ and whose sum is $b$ in the quadratic expression $x^2+bx+c.$ Then, express the factored form as $(x+m_1)(x+m_2).$
In the given expression, the value of $c$ is $
-10
$ and the value of $b$ is $
-3
.$
The possible pairs of integers whose product is $c$ are
\begin{array}{l}\require{cancel}
\{ 1,-10 \}, \{ 2,-5 \},
\{ -1,10 \}, \{ -2,5 \}
.\end{array}
Among these pairs, the one that gives a sum of $b$ is $\{
2,-5
\}.$ Hence, the factored form of the expression above is
\begin{array}{l}\require{cancel}
(x+2)(x-5)
.\end{array}