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Square root of x^2 + 3x - 10 (end square root) + 2 = x
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Sqrt[x^2 + 3x - 10] + 2 = x
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Sqrt[x^2 + 3x - 10] = x - 2; subtracted 2 from both sides
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x^2 + 3x - 10 = (x-2)^2; squared both sides
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x^2 + 3x - 10 = x^2 - 4x + 4; FOILed (x-2)^2
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x^2 - x^2 + 3x + 4x = 4 + 10; combine like terms
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7x = 14: the x^2's cancel
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x = 14/7
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x = 2
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you can check it by substitution in the original equation
