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

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

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

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

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12 - b - b^{2} = 0

b = - 4, b = 3

Solution steps

12 - b - b^{2} = 0

Write in the standard form

a x^{2} + b x + c = 0

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

Solve with the quadratic formula

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

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

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

Separate the solutions

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

b = \frac{- ( - 1 ) + 7}{2 ( - 1 )} : - 4

b = \frac{- ( - 1 ) - 7}{2 ( - 1 )} : 3

The solutions to the quadratic equation are:

b = - 4, b = 3

x (x + 1) = 110

m^{2} - 8 m + 7 = 0

(x + 3) (x + 6) = 0

3 (k^{2} - 36) + k^{2} = 6 k

x, f = x^{2} y^{3} + 3 y - 5 x

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