Answer:
[tex]\text{ From least to greatest, the solutions are x=-4 and x=-8.}[/tex]
Step-by-step explanation:
As a first step do cross multiplication and expand:
[tex]\begin{gathered} x-2=(x+10)(x+3) \\ x-2=x^2+3x+10x+30 \end{gathered}[/tex]
Organize the equation as the standard form of a quadratic equation:
[tex]ax^2+bx+c=0[/tex]
Then, organizing the equation:
[tex]\begin{gathered} x^2+12x+32=0 \\ \end{gathered}[/tex]
To solve a quadratic equation, use the quadratic formula:
[tex]x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]
Hence, for the quadratic equation above, identify a,b, and c:
a=1, b=12, and c=32.
[tex]\begin{gathered} x_1=\frac{-12+\sqrt{12^2-4(1)(32)}}{2(1)}=-\frac{8}{2}=-4 \\ x_2=\frac{-12-\sqrt{12^2-4(1)(32)}}{2(1)}=-\frac{16}{2}=-8 \end{gathered}[/tex]
