How do you solve $x^{2} + 4 x = 6$ $x^2 + 4x = 6$ by completing the square?

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Answer 1 · 2018-03-31T12:15:22

Solution: $x = - 2 + \sqrt{10} and x = - 2 - \sqrt{10}$ $x=-2+ sqrt10 and x= -2- sqrt10$

$x^{2} + 4 x = 6 or x^{2} + 4 x + 4 = 6 + 4$ $x^2+4x=6 or x^2+4x+4=6+4$ or

${(x + 2)}^{2} = 10 or x + 2 = \pm \sqrt{10}$ $(x+2)^2=10 or x+2 =+- sqrt 10$ or

$x = - 2 \pm \sqrt{10}$ $x=-2+- sqrt10$

$:.x=-2+ sqrt10 and x= -2- sqrt10$

Solution: $x=-2+ sqrt10 and x= -2- sqrt10$ [Ans]

Answer 2 · 2018-03-31T12:55:29

$x^2+4x=6$

Adding 4 on both sides.

$=>x^2+4x +4 =6 + 4$

$=>(x+2)^2 =10$

$=> x+2 = +-sqrt10$

$=> x = -2 +- sqrt10$

How do you solve x^2 + 4x = 6x2+4x=6x^2 + 4x = 6 by completing the square?