4x^2-28x=-49 how do i solve by completing the square

4x^2 -28x + 49=0
- move -49 to the other side

x^2-7x+49/4=0
- take out the 4 because you cannot complete the square with a coefficient in front of x^2

(x^2-7x +(7/2)^2) + 49/4 - (7/2)^2 = 0
- complete the square

4((x-7/2)^2)= 0
- bring the 4 back and you’re done.

hope this helped :)

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