2) \( \frac{x^{2}+2 x-3}{x+3}=-2,5 \)
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To solve the equation \( \frac{x^{2}+2 x-3}{x+3}=-2.5 \), first clear the fraction by multiplying both sides by \( x + 3 \): \( x^{2} + 2x - 3 = -2.5(x + 3) \). This simplifies to \( x^{2} + 2x - 3 = -2.5x - 7.5 \). Rearranging terms gives you \( x^{2} + 4.5x + 4.5 = 0 \). Now, you can use the quadratic formula, \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 1, b = 4.5, c = 4.5 \). Calculating the discriminant: \( b^2 - 4ac = 4.5^2 - 4 \cdot 1 \cdot 4.5 = 20.25 - 18 = 2.25 \). Since the discriminant is positive, there are two real solutions. Plugging \( a, b, \) and \( c \) into the quadratic formula gives: \[ x = \frac{-4.5 \pm 1.5}{2} \] The solutions are \( x = -1.5 \) and \( x = -3 \). However, since \( x = -3 \) would make the denominator zero, we discard it. Thus, the final solution is \( x = -1.5 \). For verification, substitute \( x = -1.5 \) back into the original equation to check if both sides equal!