Question 172185


{{{2x^2+3x-2=0}}} Start with the given equation.



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



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(2)(-2) ))/(2(2))}}} Plug in  {{{a=2}}}, {{{b=3}}}, and {{{c=-2}}}



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



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



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



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



{{{x = (-3 +- sqrt( 25 ))/(4)}}} Multiply {{{2}}} and {{{2}}} to get {{{4}}}. 



{{{x = (-3 +- 5)/(4)}}} Take the square root of {{{25}}} to get {{{5}}}. 



{{{x = (-3 + 5)/(4)}}} or {{{x = (-3 - 5)/(4)}}} Break up the expression. 



{{{x = (2)/(4)}}} or {{{x =  (-8)/(4)}}} Combine like terms. 



{{{x = 1/2}}} or {{{x = -2}}} Simplify. 



So the answers are {{{x = 1/2}}} or {{{x = -2}}}