Solve. -9x+10x^(2)+8=14 Choose 1 answer: (A) x=(3)/(4),1 (B) x=(4+-sqrt26)/(10) (c) x=(-1+-sqrt109)/(18) (D) x=(9+-sqrt321)/(20)

Rewriting the equation: First, we need to rewrite the equation in standard quadratic form, which is ax^2 + bx + c = 0. -9x + 10x^2 + 8 = 14 Subtract 14 from both sides to get: 10x^2 - 9x - 6 = 0Identifying the coefficients: Now, identify the coefficients a, b, and c from the quadratic equation 10x^2 - 9x - 6 = 0. a = 10 b = -9 c = -6Using the quadratic formula: Next, we will use the quadratic formula to find the solutions for x, which is given by: x = (-b ± sqrt(b^2 - 4ac)) / (2a)Substituting values into the formula: Substitute the values of a, b, and c into the quadratic formula. x = (-(-9) ± sqrt((-9)^2 - 4*10*(-6))) / (2*10) x = (9 ± sqrt(81 + 240)) / 20 x = (9 ± sqrt(321)) / 20Simplifying the equation: Simplify the square root and the fraction. x = (9 ± sqrt(321)) / 20 Since sqrt(321) cannot be simplified further, we leave it as is.Finding the solutions: Now we have two possible solutions for x: x = (9 + sqrt(321)) / 20 or x = (9 - sqrt(321)) / 20 These correspond to answer choice (D).

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Solve.

-9x+10x^(2)+8=14
Choose 1 answer:
(A)
x=(3)/(4),1
(B)
x=(4+-sqrt26)/(10)
(c)
x=(-1+-sqrt109)/(18)
(D)
x=(9+-sqrt321)/(20)

Solve. $\newline$ $-9 x+10 x^{2}+8=14$ $\newline$ Choose $1$ answer: $\newline$ (A) $x=\frac{3}{4}, 1$ $\newline$ (B) $x=\frac{4 \pm \sqrt{26}}{10}$ $\newline$ (C) $x=\frac{-1 \pm \sqrt{109}}{18}$ $\newline$ (D) $x=\frac{9 \pm \sqrt{321}}{20}$

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Math Problems

Algebra 2

Solve a quadratic equation using the quadratic formula

Full solution

Q. Solve.

\newline

-9 x+10 x^{2}+8=14

\newline

Choose

1

answer:

\newline

(A)

x=\frac{3}{4}, 1

\newline

(B)

x=\frac{4 \pm \sqrt{26}}{10}

\newline

(C)

x=\frac{-1 \pm \sqrt{109}}{18}

\newline

(D)

x=\frac{9 \pm \sqrt{321}}{20}

Rewriting the equation: First, we need to rewrite the equation in standard quadratic form, which is $ax^2 + bx + c = 0$ . $\newline$ $-9x + 10x^2 + 8 = 14$ $\newline$ Subtract $14$ from both sides to get: $\newline$ $10x^2 - 9x - 6 = 0$
Identifying the coefficients: Now, identify the coefficients $a$ , $b$ , and $c$ from the quadratic equation $10x^2 - 9x - 6 = 0$ . $\newline$ $a = 10$ $\newline$ $b = -9$ $\newline$ $c = -6$
Using the quadratic formula: Next, we will use the quadratic formula to find the solutions for x, which is given by: $\newline$ x = ( $-b \pm \sqrt{b^2 - 4ac}$ ) / ( $2a$ )
Substituting values into the formula: Substitute the values of $a$ , $b$ , and $c$ into the quadratic formula. $\newline$ $x = \frac{{-(-9) \pm \sqrt{{(-9)^2 - 4 \cdot 10 \cdot (-6)}}}}{{2 \cdot 10}}$ $\newline$ $x = \frac{{9 \pm \sqrt{{81 + 240}}}}{{20}}$ $\newline$ $x = \frac{{9 \pm \sqrt{{321}}}}{{20}}$
Simplifying the equation: Simplify the square root and the fraction. $\newline$ $x = \frac{9 \pm \sqrt{321}}{20}$ $\newline$ Since $\sqrt{321}$ cannot be simplified further, we leave it as is.
Finding the solutions: Now we have two possible solutions for x: $\newline$ x = $(9 + \sqrt{321}) / 20$ or x = $(9 - \sqrt{321}) / 20$ $\newline$ These correspond to answer choice $(D)$ .