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

Set Equation to Zero: First, we need to set the equation to zero by adding 1 to both sides of the equation. 2x^2 + 15x + 4 = -1 2x^2 + 15x + 4 + 1 = -1 + 1 2x^2 + 15x + 5 = 0Use Quadratic Formula: Next, we will use the quadratic formula to solve for x. The quadratic formula is x = (-b ± √(b^2 - 4ac)) / (2a), where a, b, and c are the coefficients from the quadratic equation ax^2 + bx + c = 0. In our equation, a = 2, b = 15, and c = 5.Calculate Discriminant: Now, we calculate the discriminant, which is the part under the square root in the quadratic formula: b^2 - 4ac. Discriminant = (15)^2 - 4(2)(5) Discriminant = 225 - 40 Discriminant = 185Apply Quadratic Formula: Since the discriminant is positive, we will have two real solutions. We will now apply the quadratic formula. x = (-15 ± √185) / (2 * 2) x = (-15 ± √185) / 4Calculate First Solution: We will calculate the two solutions separately. First solution: x = (-15 + √185) / 4 x ≈ (-15 + 13.601) / 4 x ≈ -1.399 / 4 x ≈ -0.34975 Rounded to the nearest tenth, x ≈ -0.3Calculate Second Solution: Second solution: x = (-15 - √185) / 4 x ≈ (-15 - 13.601) / 4 x ≈ -28.601 / 4 x ≈ -7.15025 Rounded to the nearest tenth, x ≈ -7.2

Resources

Testimonials

Plans

Resources

Testimonials

Plans

AI tutor

Welcome to Bytelearn!

Let’s check out your problem:

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

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

View step-by-step help

Home

Math Problems

Algebra 2

Solve exponential equations using common logarithms

Full solution

Q. Solve the equation

2 x^{2}+15 x+4=-1

to the nearest tenth.

\newline

Answer:

x=

Set Equation to Zero: First, we need to set the equation to zero by adding $1$ to both sides of the equation. $\newline$ $2x^2 + 15x + 4 = -1$ $\newline$ $2x^2 + 15x + 4 + 1 = -1 + 1$ $\newline$ $2x^2 + 15x + 5 = 0$
Use Quadratic Formula: Next, we will use the quadratic formula to solve for $x$ . The quadratic formula is $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ , where $a$ , $b$ , and $c$ are the coefficients from the quadratic equation $ax^2 + bx + c = 0$ . In our equation, $a = 2$ , $b = 15$ , and $c = 5$ .
Calculate Discriminant: Now, we calculate the discriminant, which is the part under the square root in the quadratic formula: $b^2 - 4ac$ .
Discriminant = $(15)^2 - 4(2)(5)$
Discriminant = $225 - 40$
Discriminant = $185$
Apply Quadratic Formula: Since the discriminant is positive, we will have two real solutions. We will now apply the quadratic formula. $\newline$ $x = \frac{-15 \pm \sqrt{185}}{2 \times 2}$ $\newline$ $x = \frac{-15 \pm \sqrt{185}}{4}$
Calculate First Solution: We will calculate the two solutions separately. $\newline$ First solution: $\newline$ $x = \frac{-15 + \sqrt{185}}{4}$ $\newline$ $x \approx \frac{-15 + 13.601}{4}$ $\newline$ $x \approx \frac{-1.399}{4}$ $\newline$ $x \approx -0.34975$ $\newline$ Rounded to the nearest tenth, $x \approx -0.3$
Calculate Second Solution: Second solution: $\newline$ $x = \frac{-15 - \sqrt{185}}{4}$ $\newline$ $x \approx \frac{-15 - 13.601}{4}$ $\newline$ $x \approx \frac{-28.601}{4}$ $\newline$ $x \approx -7.15025$ $\newline$ Rounded to the nearest tenth, $x \approx -7.2$