What are the solutions to the equation b^2-4b-32=0 ?

The solutions to the equation b^2-4b-32=0 are b=8,b=-4

Find the zeros of b^2-4b-32=0

The zeros of b^2-4b-32=0 are b=8,b=-4

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

b = 8, b = - 4

Solution steps

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} : 8

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

The solutions to the quadratic equation are:

b = 8, b = - 4

2 (3 x - 4) = 4 (\frac{1}{2} x - 8)

12 r - 12

5 x = 15

(4 - i) (4 + i)

What are the solutions to the equation b^2-4b-32=0 ?
The solutions to the equation b^2-4b-32=0 are b=8,b=-4
Find the zeros of b^2-4b-32=0
The zeros of b^2-4b-32=0 are b=8,b=-4