Question 145522
I'm assuming you want to solve for x. I'll solve the first one to get you started.



# 1





{{{x^2+7x+10=0}}} Start with the given equation.



Let's use the quadratic formula to solve for x



{{{x = (-b +- sqrt( b^2-4ac ))/(2a)}}} Start with the quadratic formula



{{{x = (-(7) +- sqrt( (7)^2-4(1)(10) ))/(2(1))}}} Plug in  {{{a=1}}}, {{{b=7}}}, and {{{c=10}}}



{{{x = (-7 +- sqrt( 49-4(1)(10) ))/(2(1))}}} Square {{{7}}} to get {{{49}}}. 



{{{x = (-7 +- sqrt( 49-40 ))/(2(1))}}} Multiply {{{4(1)(10)}}} to get {{{40}}}



{{{x = (-7 +- sqrt( 9 ))/(2(1))}}} Subtract {{{40}}} from {{{49}}} to get {{{9}}}



{{{x = (-7 +- sqrt( 9 ))/(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}. 



{{{x = (-7 +- 3)/(2)}}} Take the square root of {{{9}}} to get {{{3}}}. 



{{{x = (-7 + 3)/(2)}}} or {{{x = (-7 - 3)/(2)}}} Break up the expression. 



{{{x = (-4)/(2)}}} or {{{x =  (-10)/(2)}}} Combine like terms. 



{{{x = -2}}} or {{{x = -5}}} Simplify. 



So our answers are {{{x = -2}}} or {{{x = -5}}}