Question 170017

{{{x^2+3x=4}}} Start with the given equation.



{{{x^2+3x-4=0}}} Subtract 4 from both sides.



Notice we have a quadratic equation in the form of {{{ax^2+bx+c}}} where {{{a=1}}}, {{{b=3}}}, and {{{c=-4}}}



Let's use the quadratic formula to solve for x



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



{{{x = (-(3) +- sqrt( (3)^2-4(1)(-4) ))/(2(1))}}} Plug in  {{{a=1}}}, {{{b=3}}}, and {{{c=-4}}}



{{{x = (-3 +- sqrt( 9-4(1)(-4) ))/(2(1))}}} Square {{{3}}} to get {{{9}}}. 



{{{x = (-3 +- sqrt( 9--16 ))/(2(1))}}} Multiply {{{4(1)(-4)}}} to get {{{-16}}}



{{{x = (-3 +- sqrt( 9+16 ))/(2(1))}}} Rewrite {{{sqrt(9--16)}}} as {{{sqrt(9+16)}}}



{{{x = (-3 +- sqrt( 25 ))/(2(1))}}} Add {{{9}}} to {{{16}}} to get {{{25}}}



{{{x = (-3 +- sqrt( 25 ))/(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}. 



{{{x = (-3 +- 5)/(2)}}} Take the square root of {{{25}}} to get {{{5}}}. 



{{{x = (-3 + 5)/(2)}}} or {{{x = (-3 - 5)/(2)}}} Break up the expression. 



{{{x = (2)/(2)}}} or {{{x =  (-8)/(2)}}} Combine like terms. 



{{{x = 1}}} or {{{x = -4}}} Simplify. 



So the answers are {{{x = 1}}} or {{{x = -4}}}