Mathematics
ybb
2

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

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(2) Answers
Jayda

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

janettelove27

[latex]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 }[/latex]

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