What are the approximate solutions of 2x² + 7x + 3 rounded to the nearest hundredth?

To find the approximate solutions of the quadratic equation 2x² + 7x + 3, we can use the quadratic formula, which is given by:

x = (-b ± √(b² – 4ac)) / 2a

In this equation, a = 2, b = 7, and c = 3. Let’s substitute these values into the formula:

x = (-(7) ± √((7)² – 4(2)(3))) / (2(2))

First, we need to calculate the discriminant:

b² – 4ac = 7² – 4(2)(3) = 49 – 24 = 25

Now we can substitute the value of the discriminant back into the quadratic formula:

x = (-7 ± √25) / 4

Since the square root of 25 is 5, we have:

x = (-7 ± 5) / 4

This gives us two potential solutions:

x₁ = (-7 + 5) / 4 = -2 / 4 = -0.5

x₂ = (-7 – 5) / 4 = -12 / 4 = -3

Now, rounding these solutions to the nearest hundredth:

  • -0.5 remains -0.50
  • -3 remains -3.00

Therefore, the approximate solutions of the equation 2x² + 7x + 3 rounded to the nearest hundredth are:

-0.50 and -3.00.

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