Question

Solve by completing the square x^2+12x-28=0

Answer 1

`x^2+12x-28=0 \\ a=1\\b=12\\c=-28\\ \\ \boxed{\boxed{ \Delta=b^2-4ac}} \\ \\ \Delta=12^2-4\cdot1\cdot(-28) \\ \Delta=144+112\\ \Delta=256 \\ \\ \boxed{\boxed{\text{X}_(1,2)= (-b\pm √(\Delta) )/(2a) }} \\ \\ \\ \boxed{\text{X}_(1)= (-b - √(\Delta) )/(2a) = (-12-16)/(2) = -(28)/(2) = -14} \\ \boxed{\text{X}_2= (-b+ √(\Delta) )/(2a) = (-12+16)/(2) = (4)/(2)=2 }`

Answer 2

in ax^2+bx+c=d form

completeing the square

make sure a=1
move c to other side (by adding or subtracting so it becomes zero)
take 1/2 of b and square it so (b/2)^2
then add that o both sides
factor perfect square on left side
take square root of both sidess and remember to get positive and negative roots

1x^2+12x-28=0
a=1 good
move c to other side
add 28 to both sides
x^2+12x=28
take 1/2 of b and squaer it
12/2=6, 6^2=36
add that to both sides
x^2+12x+36=28+36
factor perfect square
(x+6)^2=64
take square root of both sides
x+6=+/-8
subtract 6 from both sides
x=-6+8 or -6-8
x=2 or -14