To solve the quadratic equation \( 2x^2 + 17x - 3 = 0 \) using the quadratic formula, we first identify the coefficients \( a \), \( b \), and \( c \):
\[ a = 2, \quad b = 17, \quad c = -3 \]
The quadratic formula is given by:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
First, we calculate the discriminant \( b^2 - 4ac \):
\[ b^2 = 17^2 = 289 \] \[ 4ac = 4 \cdot 2 \cdot (-3) = -24 \] \[ b^2 - 4ac = 289 - (-24) = 289 + 24 = 313 \]
Next, we apply the quadratic formula:
\[ x = \frac{-17 \pm \sqrt{313}}{2 \cdot 2} \]
Calculating \( \sqrt{313} \):
\[ \sqrt{313} \approx 17.69\ (rounded\ to\ two\ decimal\ places) \]
Now substituting back into the formula:
\[ x = \frac{-17 \pm 17.69}{4} \]
Now we solve for the two possible values of \( x \):
- For the plus sign:
\[ x_1 = \frac{-17 + 17.69}{4} = \frac{0.69}{4} \approx 0.1725 \approx 0.17\ (rounded\ to\ nearest\ hundredth) \]
- For the minus sign:
\[ x_2 = \frac{-17 - 17.69}{4} = \frac{-34.69}{4} \approx -8.6725 \approx -8.67\ (rounded\ to\ nearest\ hundredth) \]
Thus, the solutions to the equation \( 2x^2 + 17x - 3 = 0 \) are:
\[ \mathbf{x = 0.17} \quad \text{and} \quad \mathbf{x = -8.67} \]
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