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Solve the equation 
x^(2)+17 x+40=0 to the nearest tenth.
Answer: 
x=

Solve the equation x2+17x+40=0 x^{2}+17 x+40=0 to the nearest tenth.\newlineAnswer: x= x=

Full solution

Q. Solve the equation x2+17x+40=0 x^{2}+17 x+40=0 to the nearest tenth.\newlineAnswer: x= x=
  1. Identify Equation Type: Identify the type of equation.\newlineWe have a quadratic equation in the form ax2+bx+c=0ax^2 + bx + c = 0, where a=1a = 1, b=17b = 17, and c=40c = 40.
  2. Apply Quadratic Formula: Apply the quadratic formula to find the solutions for xx. The quadratic formula is x=b±b24ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}. Here, a=1a = 1, b=17b = 17, and c=40c = 40.
  3. Calculate Discriminant: Calculate the discriminant b24acb^2 - 4ac.\newlineDiscriminant = 1724(1)(40)=289160=12917^2 - 4(1)(40) = 289 - 160 = 129.
  4. Calculate Solutions: Calculate the two solutions using the quadratic formula.\newlinex=17±1292×1x = \frac{-17 \pm \sqrt{129}}{2 \times 1}\newlinex=17±1292x = \frac{-17 \pm \sqrt{129}}{2}
  5. Calculate First Solution: Calculate the first solution (using the positive square root).\newlinex=17+1292x = \frac{-17 + \sqrt{129}}{2}\newlinex17+11.35782x \approx \frac{-17 + 11.3578}{2}\newlinex5.64222x \approx \frac{-5.6422}{2}\newlinex2.8211x \approx -2.8211\newlineRound to the nearest tenth: x2.8x \approx -2.8
  6. Calculate Second Solution: Calculate the second solution (using the negative square root).\newlinex=171292x = \frac{{-17 - \sqrt{129}}}{{2}}\newlinex1711.35782x \approx \frac{{-17 - 11.3578}}{{2}}\newlinex28.35782x \approx \frac{{-28.3578}}{{2}}\newlinex14.1789x \approx -14.1789\newlineRound to the nearest tenth: x14.2x \approx -14.2

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