Question

(х+2) 2 -4(х+1)=5-4х

Answer 1

x^2+4x+4-4x-4-5+4x=0

x^2+4x-5=0

D= 16+20=36

x1= (-4+6)/2= 1

x2= (-4-6)/2= -5

Answer 2

Let's solve the given equation step by step:

$(x + 2)^2 - 4(x + 1) = 5 - 4x$

First, expand $(x + 2)^2$: $x^2 + 4x + 4 - 4(x + 1) = 5 - 4x$

Now distribute -4 within the parentheses: $x^2 + 4x + 4 - 4x - 4 = 5 - 4x$

Combine like terms: $x^2 + 4 - 4 = 5 - 4x$

Simplify the constants on both sides: $x^2 = 1 - 4x$

Move all terms to one side to get a quadratic equation: $x^2 + 4x - 1 = 0$

Now we have a quadratic equation in the standard form $ax^2 + bx + c = 0$, where $a = 1$, $b = 4$, and $c = -1$. We can solve this quadratic equation using various methods, such as factoring, completing the square, or using the quadratic formula.

For this equation, I'll use the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Plugging in the values $a = 1$, $b = 4$, and $c = -1$: $x = \frac{-4 \pm \sqrt{4^2 - 4(1)(-1)}}{2(1)}$ $x = \frac{-4 \pm \sqrt{16 + 4}}{2}$ $x = \frac{-4 \pm \sqrt{20}}{2}$ $x = \frac{-4 \pm 2\sqrt{5}}{2}$ $x = -2 \pm \sqrt{5}$

So, the solutions to the equation are: $x = -2 + \sqrt{5}$ $x = -2 - \sqrt{5}$

These are the two possible values for $x$ that satisfy the given equation.