Question 298951


{{{2x^2-5x-7=0}}} Start with the given equation.



Notice that the quadratic {{{2x^2-5x-7}}} is in the form of {{{Ax^2+Bx+C}}} where {{{A=2}}}, {{{B=-5}}}, and {{{C=-7}}}



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)(-7) ))/(2(2))}}} Plug in  {{{A=2}}}, {{{B=-5}}}, and {{{C=-7}}}



{{{x = (5 +- sqrt( (-5)^2-4(2)(-7) ))/(2(2))}}} Negate {{{-5}}} to get {{{5}}}. 



{{{x = (5 +- sqrt( 25-4(2)(-7) ))/(2(2))}}} Square {{{-5}}} to get {{{25}}}. 



{{{x = (5 +- sqrt( 25--56 ))/(2(2))}}} Multiply {{{4(2)(-7)}}} to get {{{-56}}}



{{{x = (5 +- sqrt( 25+56 ))/(2(2))}}} Rewrite {{{sqrt(25--56)}}} as {{{sqrt(25+56)}}}



{{{x = (5 +- sqrt( 81 ))/(2(2))}}} Add {{{25}}} to {{{56}}} to get {{{81}}}



{{{x = (5 +- sqrt( 81 ))/(4)}}} Multiply {{{2}}} and {{{2}}} to get {{{4}}}. 



{{{x = (5 +- 9)/(4)}}} Take the square root of {{{81}}} to get {{{9}}}. 



{{{x = (5 + 9)/(4)}}} or {{{x = (5 - 9)/(4)}}} Break up the expression. 



{{{x = (14)/(4)}}} or {{{x =  (-4)/(4)}}} Combine like terms. 



{{{x = 7/2}}} or {{{x = -1}}} Simplify. 



So the solutions are {{{x = 7/2}}} or {{{x = -1}}}