Solve the equation x, squared, plus, 11, x, plus, 17, equals, 0x 2 +11x+17=0 to the nearest tenth.AnswerAttempt 1 out of 2

Sure, let's solve the quadratic equation step by step.

The given equation is: $x^2 + 11x + 17 = 0$

To solve this quadratic equation, we can use the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Here, $a = 1$, $b = 11$, and $c = 17$.

1. Calculate the discriminant ($\Delta$): $\Delta = b^2 - 4ac$ $\Delta = 11^2 - 4 \cdot 1 \cdot 17$ $\Delta = 121 - 68$ $\Delta = 53$

2. Since the discriminant is positive, we have two real solutions. Now, apply the quadratic formula: $x = \frac{-b \pm \sqrt{\Delta}}{2a}$ $x = \frac{-11 \pm \sqrt{53}}{2 \cdot 1}$ $x = \frac{-11 \pm \sqrt{53}}{2}$

3. Calculate the two solutions: $x_1 = \frac{-11 + \sqrt{53}}{2}$ $x_2 = \frac{-11 - \sqrt{53}}{2}$

4. Use a calculator to find the approximate values to the nearest tenth: $\sqrt{53} \approx 7.3$

So, $x_1 = \frac{-11 + 7.3}{2} \approx \frac{-3.7}{2} \approx -1.9$ $x_2 = \frac{-11 - 7.3}{2} \approx \frac{-18.3}{2} \approx -9.2$

Therefore, the solutions to the equation $x^2 + 11x + 17 = 0$ to the nearest tenth are: $x_1 \approx -1.9$ $x_2 \approx -9.2$