Question 177461


From {{{6x^2+4x-3}}} we can see that {{{a=6}}}, {{{b=4}}}, and {{{c=-3}}}



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



{{{D=(4)^2-4(6)(-3)}}} Plug in {{{a=6}}}, {{{b=4}}}, and {{{c=-3}}}



{{{D=16-4(6)(-3)}}} Square {{{4}}} to get {{{16}}}



{{{D=16--72}}} Multiply {{{4(6)(-3)}}} to get {{{(24)(-3)=-72}}}



{{{D=16+72}}} Rewrite {{{D=16--72}}} as {{{D=16+72}}}



{{{D=88}}} Add {{{16}}} to {{{72}}} to get {{{88}}}



Since the discriminant is greater than zero, this means that the equation has two real solutions.