What are the solutions to the equation 6b^2+4b=130 ?

The solutions to the equation 6b^2+4b=130 are b= 13/3 ,b=-5

Find the zeros of 6b^2+4b=130

The zeros of 6b^2+4b=130 are b= 13/3 ,b=-5

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6 b^{2} + 4 b = 130

b = \frac{13}{3}, b = - 5

Decimal Notation

b = 4.33333 \dots, b = - 5

Solution steps

6 b^{2} + 4 b = 130

Move

130

to the left side

6 b^{2} + 4 b - 130 = 0

Solve with the quadratic formula

b_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 6 ( - 130 )}{2 \cdot 6}

4^{2} - 4 \cdot 6 (- 130) = 56

b_{1, 2} = \frac{- 4 \pm 56}{2 \cdot 6}

Separate the solutions

b_{1} = \frac{- 4 + 56}{2 \cdot 6}, b_{2} = \frac{- 4 - 56}{2 \cdot 6}

b = \frac{- 4 + 56}{2 \cdot 6} : \frac{13}{3}

b = \frac{- 4 - 56}{2 \cdot 6} : - 5

The solutions to the quadratic equation are:

b = \frac{13}{3}, b = - 5

20 k^{2} + 110 k + 120

\frac{3}{2} + \frac{2}{5} - \frac{7}{4}

3105 = 27 (p + 30)

\frac{x}{x + 5} \leq 0

9 - 2 \frac{x}{4} = - 5

What are the solutions to the equation 6b^2+4b=130 ?
The solutions to the equation 6b^2+4b=130 are b= 13/3 ,b=-5
Find the zeros of 6b^2+4b=130
The zeros of 6b^2+4b=130 are b= 13/3 ,b=-5