What are the solutions to the equation b^2+4b=21 ?

The solutions to the equation b^2+4b=21 are b=3,b=-7

Find the zeros of b^2+4b=21

The zeros of b^2+4b=21 are b=3,b=-7

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b^{2} + 4 b = 21

b = 3, b = - 7

Solution steps

b^{2} + 4 b = 21

Move

21

to the left side

b^{2} + 4 b - 21 = 0

Solve with the quadratic formula

b_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 1 \cdot ( - 21 )}{2 \cdot 1}

4^{2} - 4 \cdot 1 \cdot (- 21) = 10

b_{1, 2} = \frac{- 4 \pm 10}{2 \cdot 1}

Separate the solutions

b_{1} = \frac{- 4 + 10}{2 \cdot 1}, b_{2} = \frac{- 4 - 10}{2 \cdot 1}

b = \frac{- 4 + 10}{2 \cdot 1} : 3

b = \frac{- 4 - 10}{2 \cdot 1} : - 7

The solutions to the quadratic equation are:

b = 3, b = - 7

- q^{2} + 3 q - 2 = 0

25 x^{2} - 30 x + 13 = 0

t, x = \frac{1}{2} a t^{2}

y, x = y^{2} - y

1 - 3 x^{2} = 2 (x - 1)

What are the solutions to the equation b^2+4b=21 ?
The solutions to the equation b^2+4b=21 are b=3,b=-7
Find the zeros of b^2+4b=21
The zeros of b^2+4b=21 are b=3,b=-7