Solve the equation $2 x^{2}-15 x+20=0$ to the nearest tenth. $\newline$ Answer: $x=$

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Solve exponential equations using common logarithms

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Q. Solve the equation

2 x^{2}-15 x+20=0

to the nearest tenth.

\newline

Answer:

x=

Use Quadratic Formula: We will use the quadratic formula to solve the equation $2x^2 - 15x + 20 = 0$ . 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 case, $a = 2$ , $b = -15$ , and $c = 20$ .
Calculate Discriminant: First, we calculate the discriminant, which is the part under the square root in the quadratic formula: $b^2 - 4ac$ . Discriminant = $(-15)^2 - 4 \times 2 \times 20 = 225 - 160 = 65$ .
Plug Values and Solve: Now we can plug the values of $a$ , $b$ , and the discriminant into the quadratic formula to find the two possible values for $x$ . $\newline$ $x = \frac{-(-15) \pm \sqrt{65}}{2 \times 2}$ $\newline$ $x = \frac{15 \pm \sqrt{65}}{4}$
Calculate First Solution: We will now calculate the two possible solutions for $x$ . $\newline$ First solution: $x = \frac{15 + \sqrt{65}}{4}$ $\newline$ Second solution: $x = \frac{15 - \sqrt{65}}{4}$
Calculate Second Solution: Let's calculate the first solution: $\newline$ $x = \frac{15 + \sqrt{65}}{4}$ $\newline$ $x \approx \frac{15 + 8.062}{4}$ $\newline$ $x \approx \frac{23.062}{4}$ $\newline$ $x \approx 5.7655$ $\newline$ Rounded to the nearest tenth, $x \approx 5.8$
Calculate Second Solution: Let's calculate the first solution: $\newline$ $x = \frac{15 + \sqrt{65}}{4}$ $\newline$ $x \approx \frac{15 + 8.062}{4}$ $\newline$ $x \approx \frac{23.062}{4}$ $\newline$ $x \approx 5.7655$ $\newline$ Rounded to the nearest tenth, $x \approx 5.8$ Now, let's calculate the second solution: $\newline$ $x = \frac{15 - \sqrt{65}}{4}$ $\newline$ $x \approx \frac{15 - 8.062}{4}$ $\newline$ $x \approx \frac{6.938}{4}$ $\newline$ $x \approx 1.7345$ $\newline$ Rounded to the nearest tenth, $x \approx 1.7$