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Solve the equation:
[tex]\[ 2x^2 - 7x + 3 = 0 \][/tex]

Possible solutions:
A. [tex]\( x = -3, \quad x = -\frac{1}{2} \)[/tex]
B. [tex]\( x = 3, \quad x = \frac{1}{2} \)[/tex]

Sagot :

GinnyAnsweridn
To solve the quadratic equation:

[tex]\[2x^2 - 7x + 3 = 0\][/tex]

we can follow these steps:

### Step 1: Identify the coefficients
The general form of a quadratic equation is [tex]\(ax^2 + bx + c = 0\)[/tex]. For the given equation:

- [tex]\(a = 2\)[/tex]
- [tex]\(b = -7\)[/tex]
- [tex]\(c = 3\)[/tex]

### Step 2: Apply the quadratic formula
The quadratic formula is given by:

[tex]\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\][/tex]

### Step 3: Calculate the discriminant
The discriminant [tex]\(\Delta\)[/tex] is:

[tex]\[\Delta = b^2 - 4ac = (-7)^2 - 4 \cdot 2 \cdot 3 = 49 - 24 = 25\][/tex]

### Step 4: Find the roots
Plugging the values of [tex]\(a\)[/tex], [tex]\(b\)[/tex], and the discriminant [tex]\(\Delta\)[/tex] into the quadratic formula, we get two solutions:

[tex]\[x_{1,2} = \frac{-(-7) \pm \sqrt{25}}{2 \cdot 2} = \frac{7 \pm 5}{4}\][/tex]

These yield two results:

1. For the positive root:

[tex]\[x_1 = \frac{7 + 5}{4} = \frac{12}{4} = 3\][/tex]

2. For the negative root:

[tex]\[x_2 = \frac{7 - 5}{4} = \frac{2}{4} = \frac{1}{2}\][/tex]

### Step 5: State the solutions
Thus, the solutions to the equation [tex]\(2x^2 - 7x + 3 = 0\)[/tex] are:

[tex]\[x = \frac{1}{2} \quad \text{and} \quad x = 3\][/tex]
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