Question

Need help solving 4c^2+28c+49=0

Answer 1

Let's solve the equation; to do this we can use the general formula for quadratic equations:

`x=(-b\pm√(b^2-4ac))/(2a)`

For this equation we have a=4, b=28 and c=49; plugging these values in the general formula we have:

`\begin{gathered} c=(-28\pm√(28^2-4(4)(49)))/(2(4)) \\ =(-28\pm√(0))/(8) \\ =-(28)/(8) \\ =-(7)/(2) \end{gathered}`

Therefore, c=-7/2