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Solve the equation 
2x^(2)+15 x+4=-1 to the nearest tenth.
Answer: 
x=

Solve the equation 2x2+15x+4=1 2 x^{2}+15 x+4=-1 to the nearest tenth.\newlineAnswer: x= x=

Full solution

Q. Solve the equation 2x2+15x+4=1 2 x^{2}+15 x+4=-1 to the nearest tenth.\newlineAnswer: x= x=
  1. Set Equation to Zero: First, we need to set the equation to zero by adding 11 to both sides of the equation.\newline2x2+15x+4=12x^2 + 15x + 4 = -1\newline2x2+15x+4+1=1+12x^2 + 15x + 4 + 1 = -1 + 1\newline2x2+15x+5=02x^2 + 15x + 5 = 0
  2. Use Quadratic Formula: Next, we will use the quadratic formula to solve for xx. The quadratic formula is x=b±b24ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, where aa, bb, and cc are the coefficients from the quadratic equation ax2+bx+c=0ax^2 + bx + c = 0. In our equation, a=2a = 2, b=15b = 15, and c=5c = 5.
  3. Calculate Discriminant: Now, we calculate the discriminant, which is the part under the square root in the quadratic formula: b24acb^2 - 4ac.
    Discriminant = (15)24(2)(5)(15)^2 - 4(2)(5)
    Discriminant = 22540225 - 40
    Discriminant = 185185
  4. Apply Quadratic Formula: Since the discriminant is positive, we will have two real solutions. We will now apply the quadratic formula.\newlinex=15±1852×2x = \frac{-15 \pm \sqrt{185}}{2 \times 2}\newlinex=15±1854x = \frac{-15 \pm \sqrt{185}}{4}
  5. Calculate First Solution: We will calculate the two solutions separately.\newlineFirst solution:\newlinex=15+1854x = \frac{-15 + \sqrt{185}}{4}\newlinex15+13.6014x \approx \frac{-15 + 13.601}{4}\newlinex1.3994x \approx \frac{-1.399}{4}\newlinex0.34975x \approx -0.34975\newlineRounded to the nearest tenth, x0.3x \approx -0.3
  6. Calculate Second Solution: Second solution:\newlinex=151854x = \frac{-15 - \sqrt{185}}{4}\newlinex1513.6014x \approx \frac{-15 - 13.601}{4}\newlinex28.6014x \approx \frac{-28.601}{4}\newlinex7.15025x \approx -7.15025\newlineRounded to the nearest tenth, x7.2x \approx -7.2

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