[tex]x= \sqrt{4x+6}\\\\
x^2=4x+6\\\\
x^2-4x-6=0\\\\
a=1\\
b=-4\\
c=-6\\\\
\·=b^2-4*a*c\\
\·=(-4)^2-4*1*(-6)\\
\·=16+24\\
\·=40\\
\\
x_1= \frac{-b- \sqrt{\·} }{2a}\\\\
x_1= \frac{-(-4)- \sqrt{40} }{2*1} = \frac{4-2 \sqrt{10} }{2} = \frac{2*(2-\sqrt{10})}{2}~=\bold{ 2- \sqrt{10} }\\\\\\
x_2= \frac{-b+ \sqrt{\·} }{2a}\\\\
x_2= \frac{-(-4)+ \sqrt{40} }{2*1} = \frac{4+2 \sqrt{10} }{2} = \frac{2*(2+ \sqrt{10})}{2}~=\bold{ 2+ \sqrt{10} }\\\\\\[/tex]
x>0 - conditie de existenta
deci, luam valoarea 2+√10
