Question

5(y+3)(y-1)-4=(3+2y)в кв + y в кв

Answer 1

$\displaystyle\bf\\5(y+3)(y-1)-4=(3+2y)^{2} +y^{2} \\\\5\cdot(y^{2} -y+3y-3)-4=9+12y+4y^{2} +y^{2} \\\\5\cdot(y^{2} +2y-3)-4=9+12y+4y^{2} +y^{2} \\\\5y^{2} +10y-15-4=5y^{2} +12y+9\\\\5y^{2} +10y-5y^{2} -12y=9+15+4\\\\-2y=28\\\\\boxed{y=-14}$

Answer 2

Let's solve the equation step by step:

First, let's expand the expressions on both sides of the equation:

5(y + 3)(y - 1) - 4 = (3 + 2y)² + y²

Next, let's simplify and multiply the terms within the parentheses:

5(y² + 2y - 3) - 4 = (3 + 2y)² + y²

Now, distribute 5 to the terms inside the parentheses:

5y² + 10y - 15 - 4 = (3 + 2y)² + y²

Combine like terms on the left side:

5y² + 10y - 19 = (3 + 2y)² + y²

Next, square the binomial (3 + 2y):

5y² + 10y - 19 = (9 + 12y + 4y²) + y²

Simplify the right side by combining like terms:

5y² + 10y - 19 = 9 + 12y + 4y² + y²

Rearrange the terms by moving all the terms to one side of the equation:

5y² + 10y - 19 - 9 - 12y - 4y² - y² = 0

Combine like terms on the left side:

(5y² - 4y² + y²) + (10y - 12y) - (19 + 9) = 0

Simplify the left side:

2y² - 2y - 28 = 0

Now, the equation is in standard quadratic form (ax² + bx + c = 0), where: a = 2, b = -2, c = -28

We can solve this quadratic equation by factoring, completing the square, or using the quadratic formula. Let's solve it using the quadratic formula:

The quadratic formula states that for an equation in the form ax² + bx + c = 0, the solutions for x are given by:

x = (-b ± √(b² - 4ac)) / (2a)

Applying this formula to our equation:

x = (-(-2) ± √((-2)² - 4 * 2 * (-28))) / (2 * 2)

Simplifying further:

x = (2 ± √(4 + 224)) / 4

x = (2 ± √228) / 4

Since we're looking for real solutions, we can simplify this further:

x = (1 ± √57) / 2

Therefore, the solutions to the equation are:

x₁ = (1 + √57) / 2 x₂ = (1 - √57) / 2