Question

2(2x−13)^2−5(2x−13)+3=0

Answer 1

Відповідь:

Покрокове пояснення:

Answer 2

Ответ:

7; 7,25

Пошаговое объяснение:

Решение на фото......

Answer 3

To solve the quadratic equation $2(2x-13)^2 - 5(2x-13) + 3 = 0$, we can follow these steps:

Step 1: Let's simplify the equation by substituting $u = 2x - 13$. This substitution helps us transform the quadratic equation into a simpler form.

So the equation becomes $2u^2 - 5u + 3 = 0$.

Step 2: Now, we need to factor this quadratic equation. However, the equation doesn't seem to factor easily into simple linear factors. In this case, we can use the quadratic formula to find the solutions for $u$.

The quadratic formula is given by $u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a = 2$, $b = -5$, and $c = 3$.

Plugging in the values, we get:

$u = \frac{-(-5) \pm \sqrt{(-5)^2 - 4 \cdot 2 \cdot 3}}{2 \cdot 2}$ $u = \frac{5 \pm \sqrt{25 - 24}}{4}$ $u = \frac{5 \pm 1}{4}$

So, we have two possible values for $u$:

$u_1 = \frac{5 + 1}{4} = \frac{6}{4} = \frac{3}{2}$

$u_2 = \frac{5 - 1}{4} = \frac{4}{4} = 1$

Step 3: Recall that we substituted $u = 2x - 13$. Now we can solve for $x$ using the values of $u$:

For $u_1 = \frac{3}{2}$:

$2x - 13 = \frac{3}{2}$ $2x = \frac{3}{2} + 13$ $2x = \frac{3 + 26}{2}$ $2x = \frac{29}{2}$ $x = \frac{29}{4}$

For $u_2 = 1$:

$2x - 13 = 1$ $2x = 1 + 13$ $2x = 14$ $x = 7$

So, the solutions to the quadratic equation $2(2x-13)^2 - 5(2x-13) + 3 = 0$ are $x = \frac{29}{4}$ and $x = 7$.