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Solve the equation
6x^(2)+9x-7=-2 to the nearest tenth.
Answer:
x=

Solve the equation $6 x^{2}+9 x-7=-2$ to the nearest tenth. $\newline$ Answer: $x=$

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Solve exponential equations using common logarithms

Full solution

Q. Solve the equation

6 x^{2}+9 x-7=-2

to the nearest tenth.

\newline

Answer:

x=

Set Equation to Zero: First, we need to set the equation to zero by adding $2$ to both sides. $\newline$ $6x^2 + 9x - 7 = -2$ $\newline$ $6x^2 + 9x - 7 + 2 = 0$ $\newline$ $6x^2 + 9x - 5 = 0$
Factor or Use Quadratic Formula: Next, we will attempt to factor the quadratic equation, but if it is not factorable, we will use the quadratic formula. The quadratic formula is $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ , where $a = 6$ , $b = 9$ , and $c = -5$ .
Calculate Discriminant: Calculate the discriminant $b^2 - 4ac$ to determine if there are real solutions. $\newline$ Discriminant = $9^2 - 4(6)(-5)$ $\newline$ Discriminant = $81 + 120$ $\newline$ Discriminant = $201$ $\newline$ Since the discriminant is positive, there are two real solutions.
Apply Quadratic Formula: Now, apply the quadratic formula to find the solutions for $x$ . $x = \frac{-9 \pm \sqrt{201}}{2\cdot 6}$ $x = \frac{-9 \pm \sqrt{201}}{12}$
Calculate Solutions: Calculate the two solutions for $x$ .
First solution:
$x = \frac{-9 + \sqrt{201}}{12}$
$x \approx \frac{-9 + 14.177}{12}$
$x \approx \frac{5.177}{12}$
$x \approx 0.431$

Second solution:
$x = \frac{-9 - \sqrt{201}}{12}$
$x \approx \frac{-9 - 14.177}{12}$
$x \approx \frac{-23.177}{12}$
$x \approx -1.931$

Round both solutions to the nearest tenth.
First solution: $x \approx 0.4$
Second solution: $x = \frac{-9 + \sqrt{201}}{12}$ $0$

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