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

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

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

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

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b^{2} + b - 12 = 0

b = 3, b = - 4

Solution steps

b^{2} + b - 12 = 0

Solve with the quadratic formula

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

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

b = 3, b = - 4

10 + c = 25

\frac{2}{9} + 5

\frac{( - 5 + 6 i )}{( - 7 + 6 i )}

4 x^{2} - 4 x + 1 = 0

25 f^{2} + 10 f + 1

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