Question 300627
{{{5x^2=4x+6}}} Start with the given equation.



{{{5x^2-4x-6=0}}} Get every term to the left side.



From {{{5x^2-4x-6}}} we can see that {{{a=5}}}, {{{b=-4}}}, and {{{c=-6}}}



{{{D=b^2-4ac}}} Start with the discriminant formula.



{{{D=(-4)^2-4(5)(-6)}}} Plug in {{{a=5}}}, {{{b=-4}}}, and {{{c=-6}}}



{{{D=16-4(5)(-6)}}} Square {{{-4}}} to get {{{16}}}



{{{D=16--120}}} Multiply {{{4(5)(-6)}}} to get {{{(20)(-6)=-120}}}



{{{D=16+120}}} Rewrite {{{D=16--120}}} as {{{D=16+120}}}



{{{D=136}}} Add {{{16}}} to {{{120}}} to get {{{136}}}



Since the discriminant is greater than zero, this means that there are two real solutions.