Question 381590
Didn't first bring the 5 over to the other side.
{{{x^2+4x-5=0}}}
She could have then factored instead of using the quadratic formula to get,
{{{(x+5)(x-1)=0}}}
with {{{x=-5}}} and {{{x=1}}} as solutions.
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Instead she used {{{c=5}}} instead of {{{c=-5}}} in the quadratic formula and got the wrong answer.