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“solve the quadriatic equation 3x^2+x-5=0 to 2 decimal places” pls someone help i need the answer

User Froilan
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1 Answer

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To solve the quadratic equation 3x^2 + x - 5 = 0, we can use the quadratic formula:

x = (-b ± √(b^2 - 4ac)) / (2a)

In this equation, a, b, and c represent the coefficients of the quadratic equation.

For the given equation, the coefficients are:
a = 3
b = 1
c = -5

Substituting these values into the quadratic formula, we get:

x = (-(1) ± √((1)^2 - 4(3)(-5))) / (2(3))

Simplifying further:

x = (-1 ± √(1 + 60)) / 6
x = (-1 ± √61) / 6

Now, we can calculate the two solutions:

x₁ = (-1 + √61) / 6 ≈ 0.83 (rounded to 2 decimal places)
x₂ = (-1 - √61) / 6 ≈ -1.50 (rounded to 2 decimal places)

Therefore, the solutions to the quadratic equation 3x^2 + x - 5 = 0, rounded to 2 decimal places, are approximately:
x₁ ≈ 0.83
x₂ ≈ -1.50
User Schlangi
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