What are the solutions to the equation b^2+5b-35=3b ?

The solutions to the equation b^2+5b-35=3b are b=5,b=-7

Find the zeros of b^2+5b-35=3b

The zeros of b^2+5b-35=3b are b=5,b=-7

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b^{2} + 5 b - 35 = 3 b

b = 5, b = - 7

Solution steps

b^{2} + 5 b - 35 = 3 b

Move

3 b

to the left side

b^{2} + 2 b - 35 = 0

Solve with the quadratic formula

b_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 1 \cdot ( - 35 )}{2 \cdot 1}

2^{2} - 4 \cdot 1 \cdot (- 35) = 12

b_{1, 2} = \frac{- 2 \pm 12}{2 \cdot 1}

Separate the solutions

b_{1} = \frac{- 2 + 12}{2 \cdot 1}, b_{2} = \frac{- 2 - 12}{2 \cdot 1}

b = \frac{- 2 + 12}{2 \cdot 1} : 5

b = \frac{- 2 - 12}{2 \cdot 1} : - 7

The solutions to the quadratic equation are:

b = 5, b = - 7

\frac{\frac{1}{8}}{\frac{5}{9}} + \frac{36}{10} \cdot \frac{1}{2} - \frac{35}{100}

π \cdot 4^{2}

(x - 3)^{2} = 9

\frac{5 + x}{5 - x} \geq 1

2 (5 x - 8) = 7 - (5 - x)

What are the solutions to the equation b^2+5b-35=3b ?
The solutions to the equation b^2+5b-35=3b are b=5,b=-7
Find the zeros of b^2+5b-35=3b
The zeros of b^2+5b-35=3b are b=5,b=-7