polve the quadractic equatio using the three rell $$ 2 x^{2}+5 x^{2}+3=0 $$

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UpStudy AI · Accepted Answer

Let's first rewrite the quadratic equation you provided:

$$
2x^2 + 5x + 3 = 0
$$

Now, we can solve this quadratic equation using the quadratic formula, which is given by:

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

where $$ a $$, $$ b $$, and $$ c $$ are the coefficients from the equation $$ ax^2 + bx + c = 0 $$.

In our case:
- $$ a = 2 $$
- $$ b = 5 $$
- $$ c = 3 $$

Now, let's calculate the discriminant $$ b^2 - 4ac $$:

1. Calculate $$ b^2 $$:
   $$
   b^2 = 5^2 = 25
   $$

2. Calculate $$ 4ac $$:
   $$
   4ac = 4 \cdot 2 \cdot 3 = 24
   $$

3. Now, calculate the discriminant:
   $$
   b^2 - 4ac = 25 - 24 = 1
   $$

Since the discriminant is positive, we will have two real and distinct solutions. Now, we can substitute $$ a $$, $$ b $$, and the discriminant into the quadratic formula.

Let's calculate the two solutions:

$$
x = \frac{-5 \pm \sqrt{1}}{2 \cdot 2}
$$

This simplifies to:

$$
x = \frac{-5 \pm 1}{4}
$$

Now, we can find the two values for $$ x $$:

1. For $$ x_1 $$:
   $$
   x_1 = \frac{-5 + 1}{4} = \frac{-4}{4} = -1
   $$

2. For $$ x_2 $$:
   $$
   x_2 = \frac{-5 - 1}{4} = \frac{-6}{4} = -\frac{3}{2}
   $$

Thus, the solutions to the quadratic equation $$ 2x^2 + 5x + 3 = 0 $$ are:

$$
x_1 = -1 \quad \text{and} \quad x_2 = -\frac{3}{2}
$$
The solutions are $$ x = -1 $$ and $$ x = -\frac{3}{2} $$.

polve the quadractic equatio using the three rell \( 2 x^{2}+5 x^{2}+3=0 \)

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