2x^2+2x>4

Answer 1

x^2 + 2x > 4
add 1 to both sides
x^2 + 2x + 1 > 5
(x+1)^2 > 5

x+1>√5 or -x-1 > √5
x > √5 - 1 OR -x > √5 + 1

x > √5-1 OR x < -√5-1

Answer 2

2x^2+2x>4 Divide both sides by 2

x ^ 2 + x > 2

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take one half of thje coefficient of x and square it.

in this case ( 1 / 2 ) ^ 2 = 1 / 4
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Add 1 / 4 to both sides

x ^ 2 + x + 1 / 4 > 2 + 1 / 4

x ^ 2 + x + 1 / 4 > 8 / 4 + 1 / 4

x ^ 2 + x + 1 / 4 > 9 / 4

Factor the left hand side

( x + 1 / 2 ) ^ 2 > 9 / 4

Take the square root of both sides

abs ( x + 1 / 2 ) > 3 / 2

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abs mean absolute value
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Solutions:

1 )

x + 1 / 2 > 3 / 2 Subtract 1 / 2 to both sides

x + 1 / 2 - 1 / 2 > 3 / 2 - 1 / 2

x > 2 / 2

x > 1

2 )

- x - 1 / 2 > 3 / 2 Add 1 / 2 to both sides

- x - 1 / 2 + 1 / 2 > 3 / 2 + 1 / 2

- x > 4 / 2

- x > 2 Multiply both sides by - 1

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If you multiply or divide the inequality by a negative number
you have to flip the inequality sign.
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In this case x < - 2

So solutions are:

x < - 2 and x > 1

Answer 3

sorry Kiki

don't know how I missed that 2 in front of 2x^2

Go with Bosnian's solution

or, how about this...

2x^2 + 2x > 4
x^2 + x - 2 > 0
(x+2)(x-1) > 0

(x+2>0 and x-1<0) OR ( x+2 < 0 and x-1 > 0 )
( x>-2 and x<1) OR x< -2 and x> 1
-2 < x < 1 OR null set

-2 < x < 1