danielr_ddrp

Yes, thanks

$x-\frac{4}{-1}=x+4$

$x+4=0$

$x=-4$

$\sqrt{ {(\frac{249}{49})}^{2}}= \frac{249}{49}$

$m=\frac{\Delta Y}{\Delta X }$

$m=\frac{-6-(-7)}{3-(-3)}$

$m=\frac{1}{6}$

$\frac{1}{6} =\frac{y-(-7)}{x-(-3)}$

$\frac{x+3}{6}-7 = y$

$\frac{x-39}{6}= y$

$\sqrt{x+\sqrt{x-2}}=2 ===>> {(x+\sqrt{x-2})}^{1/2}=2 ==>>> (x+\sqrt{x-2})=4 ===>>>4-x =\sqrt{x-2}$

$4-x=\sqrt{x-2} ==>>>{(4-x)}^{2}=x-2==>>{4}^{2}-2*4*x+{x}^{2}=x-2$ 

$16-8x+{x}^{2}=x-2====>>18-9x+{x}^{2}=0$

Applying Bhaskara we gonna have x=6 or x=3

And 

$\sqrt{3+\sqrt{3-2}}=2$

X=3

$\sqrt{5^9}$