Question 335952
{{{10-5a^2=7a+9}}} Start with the given equation.



{{{10-5a^2-7a-9=0}}} Get every term to the left side.



{{{-5a^2-7a+1=0}}} Combine and rearrange the terms.



From {{{-5a^2-7a+1}}} we can see that {{{a=-5}}}, {{{b=-7}}}, and {{{c=1}}}



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



{{{D=(-7)^2-4(-5)(1)}}} Plug in {{{a=-5}}}, {{{b=-7}}}, and {{{c=1}}}



{{{D=49-4(-5)(1)}}} Square {{{-7}}} to get {{{49}}}



{{{D=49--20}}} Multiply {{{4(-5)(1)}}} to get {{{(-20)(1)=-20}}}



{{{D=49+20}}} Rewrite {{{D=49--20}}} as {{{D=49+20}}}



{{{D=69}}} Add {{{49}}} to {{{20}}} to get {{{69}}}



So the discriminant is {{{D=69}}}



Since the discriminant is greater than zero, this means that there are two real solutions. Since the discriminant is NOT a perfect square, this means that the two solutions are irrational.


So the answer is "two different irrational solutions"



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Jim