Question

(Х+3)^2+(х-2)^2=х^2-5х

Answer 1

10х-5х в кв -2-х=2х-5х в кв-6 +15=0
10х-5х в кв-2-х-2х+5х в кв+6-15х=0
-8х+4=0
х=2

Answer 2

2х+6+2х-4=-3х
4х+2=-3х
4х+3х=-2
7х=-2
Х=-2/7

Answer 3

Let's begin by expanding the left-hand side of the equation using the formula for the square of a binomial:

(Х+3)^2 = Х^2 + 6Х + 9

(x-2)^2 = x^2 - 4x + 4

(Х+3)^2+(х-2)^2 = Х^2 + 6Х + 9 + x^2 - 4x + 4 = 2x^2 + 2x + 13

Therefore, we can rewrite the equation as:

2x^2 + 2x + 13 = x^2 - 5x

Simplifying and moving all terms to one side, we get:

x^2 + 7x + 13 = 0

This is a quadratic equation. We can solve for x using the quadratic formula:

x = (-b ± sqrt(b^2 - 4ac)) / 2a

where a = 1, b = 7, and c = 13.

Plugging in the values, we get:

x = (-7 ± sqrt(7^2 - 4(1)(13))) / 2(1) x = (-7 ± sqrt(1)) / 2

x = (-7 + 1) / 2 or x = (-7 - 1) / 2

x = -3 or x = -2

Therefore, the solutions to the equation are x = -3 or x = -2.