Answer:
x = -4 and x = -16.
Step-by-step explanation:
We have the equation,
[tex]x^{2}+20x+100=36[/tex]
i.e. [tex]x^{2}+20x+64=0[/tex]
As the solution of a quadratic equation [tex]ax^{2}+bx+c=0[/tex] is given by [tex]x=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}[/tex].
On comparing, a = 1, b = 20, c = 64.
So, the solution is,
[tex]x=\frac{-20\pm \sqrt{(20)^{2}-4\times 1\times 64}}{2a\times 1}[/tex].
i.e. [tex]x=\frac{-20\pm \sqrt{400-256}}{2}[/tex].
i.e. [tex]x=\frac{-20\pm \sqrt{144}}{2}[/tex].
i.e. [tex]x=\frac{-20\pm 12}{2}[/tex].
i.e. [tex]x=\frac{-20+12}{2}[/tex] and [tex]x=\frac{-20-12}{2}[/tex].
i.e. [tex]x=\frac{-8}{2}[/tex] and [tex]x=\frac{-32}{2}[/tex].
i.e. x = -4 and x = -16
Thus, the solution of the given equation is x = -4 and x = -16.
