This problem is pretty easy if you know about discriminants.
Our equation is:
\(2x^2+5x+c=x^2-4x\)
First, move everything to the left side
\(x^2+9x+c=0\)
The discrimnant is the \(b^2-4ac\) (when the equation is \(ax^2+bx+c=0\)) part of the quadratic formula, basically
If the discriminant is greater than zero, there is 2 real solutions
If the discriminant is zero, there is one solution, a double root!
If the discriminant is less then zero, there is no real solutions
Using this we get: \(9^2-4c\) as the discrimant, which simplifies to \(81 - 4c\)
We want to get c as big as possible without this expression being negative, hence we want this to equal zero.
Solving \(81-4c=0\)
\(4c=81\)
\(c=81/4\)
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