What are the solutions to the equation b^2+7b+12=0 ?

The solutions to the equation b^2+7b+12=0 are b=-3,b=-4

Find the zeros of b^2+7b+12=0

The zeros of b^2+7b+12=0 are b=-3,b=-4

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b^{2} + 7 b + 12 = 0

b = - 3, b = - 4

Solution steps

b^{2} + 7 b + 12 = 0

Solve with the quadratic formula

b_{1, 2} = \frac{- 7 \pm 7 ^{2} - 4 \cdot 1 \cdot 12}{2 \cdot 1}

7^{2} - 4 \cdot 1 \cdot 12 = 1

b_{1, 2} = \frac{- 7 \pm 1}{2 \cdot 1}

Separate the solutions

b_{1} = \frac{- 7 + 1}{2 \cdot 1}, b_{2} = \frac{- 7 - 1}{2 \cdot 1}

b = \frac{- 7 + 1}{2 \cdot 1} : - 3

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

The solutions to the quadratic equation are:

b = - 3, b = - 4

x^{2} - 7 x + 12

4 x^{2} + 12 x + 10 = 0

- 2 x^{2} - 4 x + 5 = 0

t^{2} = 25

t^{2} = 14

What are the solutions to the equation b^2+7b+12=0 ?
The solutions to the equation b^2+7b+12=0 are b=-3,b=-4
Find the zeros of b^2+7b+12=0
The zeros of b^2+7b+12=0 are b=-3,b=-4