# How do you solve # 14x^2+3x-2=0#?

The solutions are

We can first factorise the equation and then find the solutions :

Factorising by splitting the middle term:

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To solve the quadratic equation 14x^2 + 3x - 2 = 0, you can use the quadratic formula:

x = (-b ± √(b^2 - 4ac)) / (2a)

Where a = 14, b = 3, and c = -2. Substituting these values into the formula:

x = (-3 ± √(3^2 - 4 * 14 * -2)) / (2 * 14)

x = (-3 ± √(9 + 112)) / 28

x = (-3 ± √121) / 28

x = (-3 ± 11) / 28

So the solutions are:

x₁ = (-3 + 11) / 28 = 8/28 = 2/7 x₂ = (-3 - 11) / 28 = -14/28 = -1/2

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To solve the quadratic equation (14x^2 + 3x - 2 = 0), you can use the quadratic formula, which states that for an equation in the form (ax^2 + bx + c = 0), the solutions are given by:

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

In this equation, (a = 14), (b = 3), and (c = -2). Substituting these values into the quadratic formula, we get:

[x = \frac{{-3 \pm \sqrt{{3^2 - 4 \cdot 14 \cdot (-2)}}}}{{2 \cdot 14}}]

[x = \frac{{-3 \pm \sqrt{{9 + 112}}}}{{28}}]

[x = \frac{{-3 \pm \sqrt{{121}}}}{{28}}]

[x = \frac{{-3 \pm 11}}{{28}}]

This gives two possible solutions:

[x_1 = \frac{{-3 + 11}}{{28}} = \frac{8}{28} = \frac{4}{14} = \frac{2}{7}]

[x_2 = \frac{{-3 - 11}}{{28}} = \frac{{-14}}{{28}} = -\frac{1}{2}]

So, the solutions to the equation (14x^2 + 3x - 2 = 0) are (x = \frac{2}{7}) and (x = -\frac{1}{2}).

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