3y^(2)+5y-10=0 What are the solutions to the given equation? Choose 1 answer: (A) y=-(5)/(2)-(sqrt145)/(2) and y=-(5)/(2)+(sqrt145)/(2) (B) y=-(5)/(6)-(sqrt145)/(6) and y=-(5)/(6)+(sqrt145)/(6) (c) y=(5)/(6)-(sqrt145)/(6) and y=(5)/(6)+(sqrt145)/(6) (D) y=(5)/(2)-(sqrt145)/(2) and y=(5)/(2)+(sqrt145)/(2)

Identify equation type and method: Identify the type of equation and the method to solve it. The given equation is a quadratic equation in the form ay^2 + by + c = 0, where a = 3, b = 5, and c = -10. To solve this equation, we can use the quadratic formula y = (-b ± sqrt(b^2 - 4ac)) / (2a).Apply quadratic formula: Apply the quadratic formula to find the solutions for y. Using the quadratic formula, we have: y = (-(5) ± sqrt((5)^2 - 4(3)(-10))) / (2(3)) y = (-5 ± sqrt(25 + 120)) / 6 y = (-5 ± sqrt(145)) / 6Simplify solutions: Simplify the solutions. We have two possible solutions based on the ± sign in the quadratic formula: y = (-5 + sqrt(145)) / 6 and y = (-5 - sqrt(145)) / 6Check solutions against options: Check the solutions against the given options. The solutions we found match with option (B): y = -(5)/(6) - (sqrt(145))/(6) and y = -(5)/(6) + (sqrt(145))/(6)

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3y^(2)+5y-10=0
What are the solutions to the given equation?
Choose 1 answer:
(A)
y=-(5)/(2)-(sqrt145)/(2) and

y=-(5)/(2)+(sqrt145)/(2)
(B)
y=-(5)/(6)-(sqrt145)/(6) and

y=-(5)/(6)+(sqrt145)/(6)
(c)
y=(5)/(6)-(sqrt145)/(6) and

y=(5)/(6)+(sqrt145)/(6)
(D)
y=(5)/(2)-(sqrt145)/(2) and

y=(5)/(2)+(sqrt145)/(2)

$3 y^{2}+5 y-10=0$ $\newline$ What are the solutions to the given equation? $\newline$ Choose $1$ answer: $\newline$ (A) $y=-\frac{5}{2}-\frac{\sqrt{145}}{2}$ and $\newline$ $y=-\frac{5}{2}+\frac{\sqrt{145}}{2}$ $\newline$ (B) $y=-\frac{5}{6}-\frac{\sqrt{145}}{6}$ and $\newline$ $y=-\frac{5}{6}+\frac{\sqrt{145}}{6}$ $\newline$ (C) $y=\frac{5}{6}-\frac{\sqrt{145}}{6}$ and $\newline$ $y=\frac{5}{6}+\frac{\sqrt{145}}{6}$ $\newline$ (D) $y=\frac{5}{2}-\frac{\sqrt{145}}{2}$ and $\newline$ $y=\frac{5}{2}+\frac{\sqrt{145}}{2}$

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Algebra 1

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Full solution

3 y^{2}+5 y-10=0

\newline

What are the solutions to the given equation?

\newline

Choose

1

answer:

\newline

(A)

y=-\frac{5}{2}-\frac{\sqrt{145}}{2}

and

\newline

y=-\frac{5}{2}+\frac{\sqrt{145}}{2}

\newline

(B)

y=-\frac{5}{6}-\frac{\sqrt{145}}{6}

and

\newline

y=-\frac{5}{6}+\frac{\sqrt{145}}{6}

\newline

(C)

y=\frac{5}{6}-\frac{\sqrt{145}}{6}

and

\newline

y=\frac{5}{6}+\frac{\sqrt{145}}{6}

\newline

(D)

y=\frac{5}{2}-\frac{\sqrt{145}}{2}

and

\newline

y=\frac{5}{2}+\frac{\sqrt{145}}{2}

Identify equation type and method: Identify the type of equation and the method to solve it. $\newline$ The given equation is a quadratic equation in the form $ay^2 + by + c = 0$ , where $a = 3$ , $b = 5$ , and $c = -10$ . To solve this equation, we can use the quadratic formula $y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ .
Apply quadratic formula: Apply the quadratic formula to find the solutions for $y$ . $\newline$ Using the quadratic formula, we have: $\newline$ $y = \frac{{-\left(5\right) \pm \sqrt{{\left(5\right)^2 - 4\left(3\right)\left(-10\right)}}}}{{2\left(3\right)}}$ $\newline$ $y = \frac{{-5 \pm \sqrt{25 + 120}}}{{6}}$ $\newline$ $y = \frac{{-5 \pm \sqrt{145}}}{{6}}$
Simplify solutions: Simplify the solutions. $\newline$ We have two possible solutions based on the extpm{} sign in the quadratic formula: $\newline$ y = ( $-5$ + \sqrt{ $145$ }) / $6$ and y = ( $-5$ - \sqrt{ $145$ }) / $6$
Check solutions against options: Check the solutions against the given options. $\newline$ The solutions we found match with option (B): $\newline$ y = $-\frac{5}{6} - \frac{\sqrt{145}}{6}$ and y = $-\frac{5}{6} + \frac{\sqrt{145}}{6}$