SOLUTION: this problem states to find b^2-4ac and the number of real solutions to each equation. 4m^2+25=20m this is what I did 4m^2+25-20m=0 a=4 b=-20

Question 142742This question is from textbook
: this problem states to find b^2-4ac and the number of real solutions to each equation.
4m^2+25=20m this is what I did 4m^2+25-20m=0
a=4 b=-20 c=25
(-20)^2 -4(4)(25) = 400-400= 0
Is there any other solution to this equation and if there is what do I do to find it? Thank you for your help. Anne This question is from textbook

Answer by scott8148(6628) (Show Source): You can put this solution on YOUR website!
no other solution

the 0 means the two roots (solutions) are real and equal (double root)
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