(x+4)(x+1)-3√x²+5x+2=6

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вопрос

simplify the equation:

(x+4)(x+1) - 3√x² + 5x + 2 = 6
(x^2 + 5x + 4) - 3√x^2 + 5x + 2 = 6
x^2 + 5x + 4 - 3√x^2 + 5x + 2 = 6
x^2 + 5x + 4 - 3x + 5x + 2 = 6
x^2 + 7x + 6 - 3x = 6
x^2 + 4x + 6 = 6

Now we have a quadratic equation in standard form, which is:

x^2 + 4x + 6 = 6

Subtract 6 from both sides:

x^2 + 4x = 0

Factor out an x:

x(x + 4) = 0

Now set each factor to zero and solve for x:

x = 0 or x = -4

So the solutions to the equation are x = 0 and x = -4.