# Which equation has the solutions? x^{2} + 2x + 4 = 0, x^{2} - 2x + 4 = 0, x^{2} + 2x - 4 = 0, x^{2} - 2x - 4 = 0

**Solution:**

The above quadratic equation is in the form of ax^{2 }+ bx + c

If the roots are real then the equation has the real roots, if the roots are imaginary, then the equation has no real roots.

Formula to find roots of a quadratic equation ax^{2} + bx + c = 0 is x = (-b ± √(b^{2} - 4ac))/2a

x^{2} + 2x + 4 =0

x = (-2 ± √(4 - 16))/2

= (-2 ± √-12)/2

= (-2 ± i√12)/2

Here, we have complex roots.

x^{2} - 2x + 4 =0

x = (2 ± √(4 - 16))/2

= (2 ± √-12)/2

= (2 ± i√12)/2

Here, we have complex roots.

x^{2} + 2x - 4 =0

x = (-2 ± √(4 + 16))/2

= (-2 ± √20)/2

= (-2 ± √20)/2

Here, we have real roots.

x^{2} - 2x - 4 =0

x = (2 ± √(4 + 16))/2

= (2 ± √20)/2

= (2 ± √20)/2

Here, we have real roots.

Therefore, the quadratic equations x^{2} + 2x - 4 = 0 and x^{2} - 2x - 4 = 0 have the real solutions.

## Which equation has the solutions? x^{2} + 2x + 4 = 0, x^{2} - 2x + 4 = 0, x^{2} + 2x - 4 = 0, x^{2} - 2x - 4 = 0

**Summary:**

The quadratic equations x^{2} + 2x - 4 = 0 and x^{2} - 2x - 4 = 0 have the real solutions.

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