(х2+2х)2-14(х2+2х)-15=0

To solve the equation \((x^2 + 2x)^2 - 14(x^2 + 2x) - 15 = 0\), we can start by making a substitution. Let: \[ y = x^2 + 2x \] Now, we can rewrite the equation as: \[ y^2 - 14y - 15 = 0 \] Next, we will solve for \(y\) using the quadratic formula, which is given by: \[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Here, \(a = 1\), \(b = -14\), and \(c = -15\). Let's calculate the discriminant (\(D\)): \[ D = b^2 - 4ac = (-14)^2 - 4(1)(-15) = 196 + 60 = 256 \] Since the discriminant is positive, there will be two real solutions for \(y\). Now we will apply the quadratic formula: \[ y = \frac{-(-14) \pm \sqrt{256}}{2(1)} \] \[ y = \frac{14 \pm 16}{2} \] Now calculating the two possible values for \(y\): 1. \( y_1 = \frac{14 + 16}{2} = \frac{30}{2} = 15 \) 2. \( y_2 = \frac{14 - 16}{2} = \frac{-2}{2} = -1 \) Now we have: 1. \( y = 15 \) 2. \( y = -1 \) Now we substitute back \(y = x^2 + 2x\) for both values. 1. For \(y = 15\): \[ x^2 + 2x - 15 = 0 \] Factoring this gives: \[ (x + 5)(x - 3) = 0 \] Thus, \(x = -5\) or \(x = 3\). 2. For \(y = -1\): \[ x^2 + 2x + 1 = 0 \] This can be factored as: \[ (x + 1)(x + 1) = 0 \] Thus, \(x = -1\) (a repeated root). The complete solution set for the original equation is: \[ x = -5, \quad x = 3, \quad x = -1 \] These are the solutions for the equation \((x^2 + 2x)^2 - 14(x^2 + 2x) - 15 = 0\).