Question

2y^2+8y-11 i 3y^2-5y+3

Answer 1

Ответ:

$2 {y}^{2} + 8y - 11 \\ 2 {y }^{2} + 8y - 11 = 0 \\ y = \frac{ - 4 + \sqrt{38} }{ -2} \\ y = \frac{ - 4 - \sqrt{38} }{2} \frac{ - 4 + \sqrt{38} }{2} \frac{ - 4 - \sqrt{38} }{2}$

Між двома відповідями постав кому, тому що тут помилка трохи

Answer 2

It seems like you have provided two quadratic expressions. Let's take a look at each of them separately:

Expression 1: 2y^2 + 8y - 11

Expression 2: 3y^2 - 5y + 3

These are quadratic equations in terms of the variable 'y'. To solve them, we can use various methods such as factoring, completing the square, or using the quadratic formula.

Let's solve each equation separately:

Expression 1: 2y^2 + 8y - 11

To solve this equation, we can use the quadratic formula:

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

x = (-b ± sqrt(b^2 - 4ac)) / (2a)

In our case, a = 2, b = 8, and c = -11. Plugging these values into the quadratic formula, we get:

y = (-8 ± sqrt(8^2 - 4 * 2 * -11)) / (2 * 2)

Simplifying further:

y = (-8 ± sqrt(64 + 88)) / 4

y = (-8 ± sqrt(152)) / 4

y = (-8 ± sqrt(4 * 38)) / 4

y = (-8 ± 2sqrt(38)) / 4

Now, we can simplify the expression:

y = (-4 ± sqrt(38)) / 2

Therefore, the solutions for the equation 2y^2 + 8y - 11 = 0 are:

y = (-4 + sqrt(38)) / 2

y = (-4 - sqrt(38)) / 2

Now let's move on to the second expression:

Expression 2: 3y^2 - 5y + 3

To solve this equation, we can again use the quadratic formula:

In this case, a = 3, b = -5, and c = 3. Plugging these values into the quadratic formula, we get:

y = (-(-5) ± sqrt((-5)^2 - 4 * 3 * 3)) / (2 * 3)

Simplifying further:

y = (5 ± sqrt(25 - 36)) / 6

y = (5 ± sqrt(-11)) / 6

Since the discriminant is negative, the solutions will be complex numbers. Simplifying further:

y = (5 ± sqrt(11)i) / 6

Therefore, the solutions for the equation 3y^2 - 5y + 3 = 0 are:

y = (5 + sqrt(11)i) / 6

y = (5 - sqrt(11)i) / 6

Note: The 'i' represents the imaginary unit, where 'i' is defined as the square root of -1.