Question

4x^2-28x+49=5 solve the quadratic by completing the square

Answer 1

`4x^2-28x+49=5 \\ \\4x^2-28x+49-5=0\\ \\4x^2-28x+44=0\\ \\ \Delta = b^(2)-4ac =(-28)^(2)-4*4*44=784 - 704 = 80 \\ \\x_(1)=(-b-√(\Delta ))/(2a) =(28- √(80))/(2*4)=(-28-√(16*5))/(8)= (-28-4√( 5))/(8)=\\ \\=(\\ot4^1(-7- √( 5)))/(\\ot8^2)=( -7- √( 5) )/(2)=`


`x_(2)=(-b+√(\Delta ))/(2a) =(28+√(80))/(2*4)=(-28+√(16*5))/(8)= (-28+4√( 5))/(8)=\\ \\=(\\ot4^1(-7+ √( 5)))/(\\ot8^2)=( -7+ √( 5) )/(2) \\ \\Answer : x=( -7- √( 5) )/(2) \ \ or \ \ x =( -7+ √( 5) )/(2)`