What are the solutions to the equation b^2+5b=32+b ?

The solutions to the equation b^2+5b=32+b are b=4,b=-8

Find the zeros of b^2+5b=32+b

The zeros of b^2+5b=32+b are b=4,b=-8

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b^{2} + 5 b = 32 + b

b = 4, b = - 8

Solution steps

b^{2} + 5 b = 32 + b

Move

b

to the left side

b^{2} + 4 b = 32

Move

32

to the left side

b^{2} + 4 b - 32 = 0

Solve with the quadratic formula

b_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 1 \cdot ( - 32 )}{2 \cdot 1}

4^{2} - 4 \cdot 1 \cdot (- 32) = 12

b_{1, 2} = \frac{- 4 \pm 12}{2 \cdot 1}

Separate the solutions

b_{1} = \frac{- 4 + 12}{2 \cdot 1}, b_{2} = \frac{- 4 - 12}{2 \cdot 1}

b = \frac{- 4 + 12}{2 \cdot 1} : 4

b = \frac{- 4 - 12}{2 \cdot 1} : - 8

The solutions to the quadratic equation are:

b = 4, b = - 8

(\frac{1}{7})^{2} + y^{2} = 1

2.7 \cdot 1 0^{- 9} x^{2} - 16 \cdot 1 0^{- 5} x + 0.5 = 0

x, x^{2} - x y = 23

y^{2} - 8 y = 0

x, (x + k) (x^{2} + 3 x - 1) = x^{3} - x^{2} - 13 x + 4

What are the solutions to the equation b^2+5b=32+b ?
The solutions to the equation b^2+5b=32+b are b=4,b=-8
Find the zeros of b^2+5b=32+b
The zeros of b^2+5b=32+b are b=4,b=-8