Question 179510
# 1




{{{2x^2=40}}} Start with the given equation.



{{{x^2=(40)/(2)}}} Divide both sides by {{{2}}}.



{{{x^2=20}}} Reduce.



{{{x=0+-sqrt(20)}}} Take the square root of both sides.



{{{x=sqrt(20)}}} or {{{x=-sqrt(20)}}} Break up the "plus/minus" to form two equations.



{{{x=2*sqrt(5)}}} or {{{x=-2*sqrt(5)}}}  Simplify the square root.



--------------------------------------



Answer:



So the solutions are {{{x=2*sqrt(5)}}} or {{{x=-2*sqrt(5)}}}.




<hr>



# 2





{{{(x+3)^2=49}}} Start with the given equation.



{{{x+3=0+-sqrt(49)}}} Take the square root of both sides.



{{{x+3=sqrt(49)}}} or {{{x+3=-sqrt(49)}}} Break up the "plus/minus" to form two equations.



{{{x+3=7}}} or {{{x+3=-7}}}  Take the square root of {{{49}}} to get {{{7}}}.



{{{x=-3+7}}} or {{{x=-3-7}}} Subtract {{{3}}} from both sides.



{{{x=4}}} or {{{x=-10}}} Combine like terms.



--------------------------------------



Answer:



So the solutions are {{{x=4}}} or {{{x=-10}}}.