What are the solutions to the equation 3b^2-18b=10b-49 ?

The solutions to the equation 3b^2-18b=10b-49 are b=7,b= 7/3

Find the zeros of 3b^2-18b=10b-49

The zeros of 3b^2-18b=10b-49 are b=7,b= 7/3

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3 b^{2} - 18 b = 10 b - 49

b = 7, b = \frac{7}{3}

Solution steps

3 b^{2} - 18 b = 10 b - 49

Move

49

to the left side

3 b^{2} - 18 b + 49 = 10 b

Move

10 b

to the left side

3 b^{2} - 28 b + 49 = 0

Solve with the quadratic formula

b_{1, 2} = \frac{- ( - 28 ) \pm ( - 28 ) ^{2} - 4 \cdot 3 \cdot 49}{2 \cdot 3}

(- 28)^{2} - 4 \cdot 3 \cdot 49 = 14

b_{1, 2} = \frac{- ( - 28 ) \pm 14}{2 \cdot 3}

Separate the solutions

b_{1} = \frac{- ( - 28 ) + 14}{2 \cdot 3}, b_{2} = \frac{- ( - 28 ) - 14}{2 \cdot 3}

b = \frac{- ( - 28 ) + 14}{2 \cdot 3} : 7

b = \frac{- ( - 28 ) - 14}{2 \cdot 3} : \frac{7}{3}

The solutions to the quadratic equation are:

b = 7, b = \frac{7}{3}

- \frac{5}{2} - 2

9 x - 3 > 6

\frac{2}{3} r + 1 = 23

(x^{- 1} + 2^{- 1})^{- 1}

x^{2} - 9 x + 14 > 0

What are the solutions to the equation 3b^2-18b=10b-49 ?
The solutions to the equation 3b^2-18b=10b-49 are b=7,b= 7/3
Find the zeros of 3b^2-18b=10b-49
The zeros of 3b^2-18b=10b-49 are b=7,b= 7/3