Rewrite in standard form: Start by rewriting the equation in standard quadratic form. We want to get the equation into the form ax^2 + bx + c = 0. To do this, we move all terms to one side of the equation: 10 = 17b - 6b^2 6b^2 - 17b + 10 = 0Factor the quadratic equation: Factor the quadratic equation. We look for two numbers that multiply to 6*10=60 and add up to -17. These numbers are -15 and -2. So we can write the equation as: (6b - 2)(b - 5) = 0Solve for b: Solve for b using the zero product property. If the product of two factors is zero, then at least one of the factors must be zero. Therefore, we set each factor equal to zero and solve for b: 6b - 2 = 0 or b - 5 = 0Solve first equation for b: Solve the first equation 6b - 2 = 0 for b. Add 2 to both sides: 6b = 2 Divide both sides by 6: b = 2/6 Simplify the fraction: b = 1/3Solve second equation for b: Solve the second equation b - 5 = 0 for b. Add 5 to both sides: b = 5Check solutions: Check the solutions in the original equation. For b = 1/3: 10 = 17(1/3) - 6(1/3)^2 10 = 17/3 - 6/9 10 = 51/3 - 2/3 10 = 49/3 10 ≠ 49/3, so there is a math error.

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$10=17 b-6 b^{2}$

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Algebra 2

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10=17 b-6 b^{2}

Rewrite in standard form: Start by rewriting the equation in standard quadratic form. $\newline$ We want to get the equation into the form $ax^2 + bx + c = 0$ . To do this, we move all terms to one side of the equation: $\newline$ $10 = 17b - 6b^2$ $\newline$ $6b^2 - 17b + 10 = 0$
Factor the quadratic equation: Factor the quadratic equation. $\newline$ We look for two numbers that multiply to $6\times10=60$ and add up to $-17$ . These numbers are $-15$ and $-2$ . $\newline$ So we can write the equation as: $\newline$ $(6b - 2)(b - 5) = 0$
Solve for b: Solve for b using the zero product property. $\newline$ If the product of two factors is zero, then at least one of the factors must be zero. Therefore, we set each factor equal to zero and solve for b: $\newline$ $6b - 2 = 0$ or $b - 5 = 0$
Solve first equation for $b$ : Solve the first equation $6b - 2 = 0$ for $b$ . Add $2$ to both sides: $6b = 2$ Divide both sides by $6$ : $b = \frac{2}{6}$ Simplify the fraction: $b = \frac{1}{3}$
Solve second equation for $b$ : Solve the second equation $b - 5 = 0$ for $b$ . $\newline$ Add $5$ to both sides: $\newline$ $b = 5$
Check solutions: Check the solutions in the original equation. $\newline$ For $b = \frac{1}{3}$ : $\newline$ $10 = 17\left(\frac{1}{3}\right) - 6\left(\frac{1}{3}\right)^2$ $\newline$ $10 = \frac{17}{3} - \frac{6}{9}$ $\newline$ $10 = \frac{51}{3} - \frac{2}{3}$ $\newline$ $10 = \frac{49}{3}$ $\newline$ $10 \neq \frac{49}{3}$ , so there is a math error.