1. You can check whether 4 and -6 are solutions by substituting each of them in turn into the original equations. Since 4^2 + 2.4 - 24 = 0 and (-6)^2 + 2.(-6) + 24 = 0 it turns out that they are indeed solutions. Sine a quadratic equations has at most two roots, that's the complete set.

2. If x^2 - 25 = 0 then x^2 = 25. Taking square roots, x = +5 or -5.

3. 2x + 1/7 = x + 2/5. Subtract x from both sides: x + 1/7 = 2/5. Subtract 1/7 from both sides: x = 2/5 - 1/7. Express RHS in terms of a common denominator 5.7 = 35: x = 14/35 - 5/35 = 9/35.

4. x-2/x + 1/2 = x+2/2x. Multiply through by x: x^2 - 2 + x/2 = x^2 + 2/2. Simplifying, x/2 = 3: x = 6.

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6. x^2 + 5x - 14. To complete the square, firstly divide through by the coefficient of x^2. In this case, that's already done. Now take half the coefficient of x, here 5/2 and add and subtract it: x^2 + 5x + (5/2)^2 - (5/2)^2 - 14. This makes the first part into an exact square (x+5/2)^2 - (5/2)^2 - 14. Now (x+5/2)^2 = 14 + 25/4 = 169/4. Taking square roots, x+5/2 = + or - 13/2. So x = 8/2=4 or -18/2 = 9. Remember to substitute into the original to ckeck!

7. x^2 -5x -24 by the formula. a = coefficient of x^2 = 1, b = coefficient of x = -5, c = constant term = -24. Formula: x = -b +or- sqrt(b^2-4ac) all over 2a.

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