Question 194496


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



Notice we have a quadratic equation in the form of {{{ax^2+bx+c}}} where {{{a=5}}}, {{{b=2}}}, 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 = (-(2) +- sqrt( (2)^2-4(5)(-3) ))/(2(5))}}} Plug in  {{{a=5}}}, {{{b=2}}}, and {{{c=-3}}}



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



{{{x = (-2 +- sqrt( 4--60 ))/(2(5))}}} Multiply {{{4(5)(-3)}}} to get {{{-60}}}



{{{x = (-2 +- sqrt( 4+60 ))/(2(5))}}} Rewrite {{{sqrt(4--60)}}} as {{{sqrt(4+60)}}}



{{{x = (-2 +- sqrt( 64 ))/(2(5))}}} Add {{{4}}} to {{{60}}} to get {{{64}}}



{{{x = (-2 +- sqrt( 64 ))/(10)}}} Multiply {{{2}}} and {{{5}}} to get {{{10}}}. 



{{{x = (-2 +- 8)/(10)}}} Take the square root of {{{64}}} to get {{{8}}}. 



{{{x = (-2 + 8)/(10)}}} or {{{x = (-2 - 8)/(10)}}} Break up the expression. 



{{{x = (6)/(10)}}} or {{{x =  (-10)/(10)}}} Combine like terms. 



{{{x = 3/5}}} or {{{x = -1}}} Simplify. 



So the answers are {{{x = 3/5}}} or {{{x = -1}}}