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Solve the equation 
3x^(2)-5x-5=0 to the nearest tenth.
Answer: 
x=

Solve the equation 3x25x5=0 3 x^{2}-5 x-5=0 to the nearest tenth.\newlineAnswer: x= x=

Full solution

Q. Solve the equation 3x25x5=0 3 x^{2}-5 x-5=0 to the nearest tenth.\newlineAnswer: x= x=
  1. Calculate Discriminant: To solve the quadratic equation 3x25x5=03x^{2}-5x-5=0, we can use the quadratic formula, which is x=b±b24ac2ax = \frac{-b \pm \sqrt{b^2-4ac}}{2a}, where aa, bb, and cc are the coefficients from the equation ax2+bx+c=0ax^2 + bx + c = 0. In our case, a=3a = 3, b=5b = -5, and c=5c = -5.
  2. Plug into Quadratic Formula: First, we calculate the discriminant, which is the part under the square root in the quadratic formula: b24acb^2 - 4ac.\newlineDiscriminant = (5)24×3×(5)=25+60=85(-5)^2 - 4 \times 3 \times (-5) = 25 + 60 = 85.
  3. Calculate Solutions: Now we can plug the discriminant and the coefficients into the quadratic formula to find the two possible values for xx.x=(5)±852×3x = \frac{-(-5) \pm \sqrt{85}}{2 \times 3}x=5±856x = \frac{5 \pm \sqrt{85}}{6}
  4. Approximate Square Root: We will now calculate the two possible solutions for xx.\newlineFirst solution: x=5+856x = \frac{5 + \sqrt{85}}{6}\newlineSecond solution: x=5856x = \frac{5 - \sqrt{85}}{6}
  5. Round Solutions: Using a calculator, we find the approximate values for the square root of 8585 and then for xx.859.220\sqrt{85} \approx 9.220First solution: x(5+9.220)/614.220/62.370x \approx (5 + 9.220) / 6 \approx 14.220 / 6 \approx 2.370Second solution: x(59.220)/64.220/60.703x \approx (5 - 9.220) / 6 \approx -4.220 / 6 \approx -0.703
  6. Round Solutions: Using a calculator, we find the approximate values for the square root of 8585 and then for x.\newline859.220\sqrt{85} \approx 9.220\newlineFirst solution: x(5+9.220)/614.220/62.370x \approx (5 + 9.220) / 6 \approx 14.220 / 6 \approx 2.370\newlineSecond solution: x(59.220)/64.220/60.703x \approx (5 - 9.220) / 6 \approx -4.220 / 6 \approx -0.703Finally, we round the solutions to the nearest tenth.\newlineFirst solution: x2.4x \approx 2.4\newlineSecond solution: x0.7x \approx -0.7

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