Suppose that a is a nonzero constant for which the equation ax^2+20x+7=0 has only one solution. find this solution.

1 answer:

4 0

$\Delta=0\\\\
\Delta=20^2-4\cdot a\cdot7=400-28a\\
400-28a=0\\
28a=400\\
a=\dfrac{400}{28}=\dfrac{100}{7}

$

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