Решите уравнения 1.sqrt(3x+7)=x+3 2.x=5-sqrt(2x^2+13-14x)

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Square both sides:
3x + 7 = (x + 3)^2
3x + 7 = x^2 + 6x + 9

Rearranging and solving for x:
x^2 + 3x - 2 = 0
(x + 2)(x - 1) = 0

x + 2 = 0 or x - 1 = 0
x = -2 or x = 1

So, the solutions are x = -2 or x = 1.

To solve this equation, we first isolate the square root term on one side:
Add sqrt(2x^2 + 13 - 14x) to both sides:
x + sqrt(2x^2 + 13 - 14x) = 5

Square both sides to get rid of the square root:
(x + sqrt(2x^2 + 13 - 14x))^2 = 5^2
(x + sqrt(2x^2 + 13 - 14x))(x + sqrt(2x^2 + 13 - 14x)) = 25

Expanding the left side:
x^2 + xsqrt(2x^2 + 13 - 14x) + xsqrt(2x^2 + 13 - 14x) + 2x^2 + 13 - 14x = 25
x^2 + 2xsqrt(2x^2 + 13 - 14x) + 3x^2 + 13 - 14x = 25
3x^2 + 2xsqrt(2x^2 + 13 - 14x) - 14x + 13 = 25
3x^2 + 2x*sqrt(2x^2 + 13 - 14x) = 12

Now, square both sides again to eliminate the square root term:
(3x^2 + 2xsqrt(2x^2 + 13 - 14x))^2 = 12^2
9x^4 + 12x^3sqrt(2x^2 + 13 - 14x) + 4x^2(2x^2 + 13 - 14x) = 144
9x^4 + 12x^3*sqrt(2x^2 + 13 - 14x) + 8x^4 + 52x^2 - 56x^3 = 144
17x^4 - 44x^3 + 52x^2 - 144 = 0

This is a quartic equation that can be solved using numerical methods as it does not have an easy algebraic solution.