Answer:
The roots are [tex]x=\dfrac{-7+ \sqrt{97}}{2}\text{ and }x=\dfrac{-7- \sqrt{97}}{2}[/tex].
Step-by-step explanation:
Consider the provided equation.
[tex]2(6 - x) = x(x + 5)[/tex]
Remove the parenthesis and simplify.
[tex]12-2 x= x^2 + 5x[/tex]
[tex]x^2+7x-12=0[/tex]
Now use the quadratic formula [tex]x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex].
Susbtitute a=1, b=7 and c=-12 in above formula.
[tex]x=\dfrac{-7\pm \sqrt{7^2-4(1)(-12)}}{2(1)}[/tex]
[tex]x=\dfrac{-7\pm \sqrt{49+48}}{2}[/tex]
[tex]x=\dfrac{-7\pm \sqrt{97}}{2}[/tex]
Hence, the roots are [tex]x=\dfrac{-7+ \sqrt{97}}{2}\text{ and }x=\dfrac{-7- \sqrt{97}}{2}[/tex].