Question

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

Answer 1

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

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