Question

(x+7)(x-2)+(x+4)(x+3)=-2

Answer 1

Ответ:

${x}^{2} + 7x - 2x - 14 + {x}^{2} + 4x + 3x + 12 = - 2 \\ 2 {x}^{2} + 12x = 0 \\ 2x (x + 6) = 0 \\ x = 0 \\ x = - 6$

Answer 2

$(x+7)(x-2)+(x+4)(x+3)=-2\\\\x^{2}-2x+7x-14+x^{2}+3x+4x+12=-2\\\\2x^{2}+12x-2=-2\\\\2x^{2} +12x=0\\\\x^{2}+6x=0\\\\x(x+6)=0\\\\\left[\begin{array}{ccc}x=0\\x+6=0\end{array}\right\\\\\\\left[\begin{array}{ccc}x_{1}=0 \\x_{2}=-6 \end{array}\right\\\\Otvet:\boxed{-6;0}$

Answer 3

Let's solve the given equation step by step.

Expanding the equation: (x + 7)(x - 2) + (x + 4)(x + 3) = -2

Using the distributive property: (x * x) + (x * -2) + (7 * x) + (7 * -2) + (x * x) + (x * 3) + (4 * x) + (4 * 3) = -2

Simplifying: x^2 - 2x + 7x - 14 + x^2 + 3x + 4x + 12 = -2

Combining like terms: 2x^2 + 12x - 2 = -2

Subtracting -2 from both sides: 2x^2 + 12x = 0

Dividing both sides by 2 to simplify: x^2 + 6x = 0

Factoring out an x: x(x + 6) = 0

Now, we have two possible solutions:

1. x = 0
2. x + 6 = 0 x = -6

So, the solutions to the equation (x + 7)(x - 2) + (x + 4)(x + 3) = -2 are x = 0 and x = -6.