Question

(2x2 + 6x – 7) + (A – 5x) = 3x2 – 7x – 1​

Answer 1

Ответ:

x2-8x+6

Объяснение:

тексердім =)

Answer 2

To solve the given equation, let's first simplify both sides:

Starting with the left-hand side (LHS): (2x^2 + 6x - 7) + (A - 5x)

Now let's combine like terms: 2x^2 + 6x - 7 + A - 5x

Next, let's simplify the right-hand side (RHS): 3x^2 - 7x - 1

Now we can set the LHS equal to the RHS and solve for x: 2x^2 + 6x - 7 + A - 5x = 3x^2 - 7x - 1

Rearranging the terms: 2x^2 - 3x^2 + 6x + 7x + 5x + 7 + 1 = 0

Simplifying further: -x^2 + 18x + 8 = 0

To solve this quadratic equation, we can use the quadratic formula: x = (-b ± sqrt(b^2 - 4ac)) / (2a)

Comparing the equation to the standard quadratic form ax^2 + bx + c = 0, we have: a = -1, b = 18, c = 8

Substituting the values into the quadratic formula: x = (-18 ± sqrt(18^2 - 4(-1)(8))) / (2(-1))

Simplifying: x = (-18 ± sqrt(324 + 32)) / (-2) x = (-18 ± sqrt(356)) / (-2) x = (-18 ± sqrt(89 * 4)) / (-2) x = (-18 ± 2sqrt(89)) / (-2) x = 9 ± sqrt(89)

Therefore, the solutions to the equation are: x = 9 + sqrt(89) x = 9 - sqrt(89)