Question 175636


{{{2x^2+5x-3=0}}} Start with the given equation.



Notice we have a quadratic equation in the form of {{{ax^2+bx+c}}} where {{{a=2}}}, {{{b=5}}}, and {{{c=-3}}}



Let's use the quadratic formula to solve for x



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



{{{x = (-(5) +- sqrt( (5)^2-4(2)(-3) ))/(2(2))}}} Plug in  {{{a=2}}}, {{{b=5}}}, and {{{c=-3}}}



{{{x = (-5 +- sqrt( 25-4(2)(-3) ))/(2(2))}}} Square {{{5}}} to get {{{25}}}. 



{{{x = (-5 +- sqrt( 25--24 ))/(2(2))}}} Multiply {{{4(2)(-3)}}} to get {{{-24}}}



{{{x = (-5 +- sqrt( 25+24 ))/(2(2))}}} Rewrite {{{sqrt(25--24)}}} as {{{sqrt(25+24)}}}



{{{x = (-5 +- sqrt( 49 ))/(2(2))}}} Add {{{25}}} to {{{24}}} to get {{{49}}}



{{{x = (-5 +- sqrt( 49 ))/(4)}}} Multiply {{{2}}} and {{{2}}} to get {{{4}}}. 



{{{x = (-5 +- 7)/(4)}}} Take the square root of {{{49}}} to get {{{7}}}. 



{{{x = (-5 + 7)/(4)}}} or {{{x = (-5 - 7)/(4)}}} Break up the expression. 



{{{x = (2)/(4)}}} or {{{x =  (-12)/(4)}}} Combine like terms. 



{{{x = 1/2}}} or {{{x = -3}}} Simplify. 



So the answers are {{{x = 1/2}}} or {{{x = -3}}}