Question

(x-4)(x-5)+(3-x)(x-2)=0

Answer 1

X²-5x-4x+20+3x-6-x²+2x=0
-9x+5x=-20+6
-4x=-14
x=14/4
x=3,5

Answer 2

X^2 - 5x -4x + 20 + 3x - 6 - x^2 + 2x = 0
-9x + 14 + 5x = 0
-4x = - 14 
x = 14/4 
x = 3,5

Answer 3

To solve the quadratic equation $(x-4)(x-5) + (3-x)(x-2) = 0$, you'll need to expand the expressions, simplify, and then solve for $x$.

Let's start by expanding the expressions:

$(x-4)(x-5) + (3-x)(x-2) = 0$

Expand each term:

$x^2 - 5x - 4x + 20 + 3x - 2x - 6 + 2x = 0$

Combine like terms:

$x^2 - 6x + 20 - 6 + 2x = 0$

Simplify:

$x^2 - 4x + 14 = 0$

Now, you have a quadratic equation in standard form:

$x^2 - 4x + 14 = 0$

To solve for $x$, you can use the quadratic formula:

The quadratic formula states that for an equation of the form $ax^2 + bx + c = 0$, the solutions for $x$ are given by:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

In this case, $a = 1$, $b = -4$, and $c = 14$.

Plugging in these values:

$x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4 \cdot 1 \cdot 14}}{2 \cdot 1}$

$x = \frac{4 \pm \sqrt{16 - 56}}{2}$

$x = \frac{4 \pm \sqrt{-40}}{2}$

Since the square root of a negative number is not a real number, this equation has complex solutions:

$x = \frac{4 \pm \sqrt{40}i}{2}$

Simplifying further:

$x = 2 \pm \sqrt{10}i$

So, the solutions for the equation $(x-4)(x-5) + (3-x)(x-2) = 0$ are $x = 2 + \sqrt{10}i$ and $x = 2 - \sqrt{10}i$, where $i$ is the imaginary unit.