Question 144184
{{{4m^2+25=20m}}} Start with the given equation



{{{4m^2+25-20m=0}}}  Subtract 20m from both sides. 



{{{4m^2-20m+25=0}}}  Rearrange the terms 



Let's use the quadratic formula to solve for m:



Starting with the general quadratic


{{{am^2+bm+c=0}}}


the general solution using the quadratic equation is:


{{{m = (-b +- sqrt( b^2-4*a*c ))/(2*a)}}}




So lets solve {{{4*m^2-20*m+25=0}}} ( notice {{{a=4}}}, {{{b=-20}}}, and {{{c=25}}})





{{{m = (--20 +- sqrt( (-20)^2-4*4*25 ))/(2*4)}}} Plug in a=4, b=-20, and c=25




{{{m = (20 +- sqrt( (-20)^2-4*4*25 ))/(2*4)}}} Negate -20 to get 20




{{{m = (20 +- sqrt( 400-4*4*25 ))/(2*4)}}} Square -20 to get 400  (note: remember when you square -20, you must square the negative as well. This is because {{{(-20)^2=-20*-20=400}}}.)




{{{m = (20 +- sqrt( 400+-400 ))/(2*4)}}} Multiply {{{-4*25*4}}} to get {{{-400}}}




{{{m = (20 +- sqrt( 0 ))/(2*4)}}} Combine like terms in the radicand (everything under the square root)




{{{m = (20 +- 0)/(2*4)}}} Simplify the square root 




{{{m = (20 +- 0)/8}}} Multiply 2 and 4 to get 8


So now the expression breaks down into two parts


{{{m = (20 + 0)/8}}} or {{{m = (20 - 0)/8}}}


Lets look at the first part:


{{{x=(20 + 0)/8}}}


{{{m=20/8}}} Add the terms in the numerator

{{{m=5/2}}} Divide


So one answer is

{{{m=5/2}}}




Now lets look at the second part:


{{{x=(20 - 0)/8}}}


{{{m=20/8}}} Subtract the terms in the numerator

{{{m=5/2}}} Divide


So another answer is

{{{m=5/2}}}




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Answer:


So the only solution is


{{{m=5/2}}}