What are the solutions to the equation b^2-4b+4=16 ?

The solutions to the equation b^2-4b+4=16 are b=6,b=-2

Find the zeros of b^2-4b+4=16

The zeros of b^2-4b+4=16 are b=6,b=-2

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b^{2} - 4 b + 4 = 16

b = 6, b = - 2

Solution steps

b^{2} - 4 b + 4 = 16

Move

16

to the left side

b^{2} - 4 b - 12 = 0

Solve with the quadratic formula

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

(- 4)^{2} - 4 \cdot 1 \cdot (- 12) = 8

b_{1, 2} = \frac{- ( - 4 ) \pm 8}{2 \cdot 1}

Separate the solutions

b_{1} = \frac{- ( - 4 ) + 8}{2 \cdot 1}, b_{2} = \frac{- ( - 4 ) - 8}{2 \cdot 1}

b = \frac{- ( - 4 ) + 8}{2 \cdot 1} : 6

b = \frac{- ( - 4 ) - 8}{2 \cdot 1} : - 2

The solutions to the quadratic equation are:

b = 6, b = - 2

r^{2} + 6 r + \frac{279}{4} = 0

(x - 3) (x + 18) = 0

\frac{x ^{2}}{8 + 3} = 19

12.5 x^{2} + 92.4 x - 4158 = 0

\frac{x ^{2}}{4} = 2 x - 3

What are the solutions to the equation b^2-4b+4=16 ?
The solutions to the equation b^2-4b+4=16 are b=6,b=-2
Find the zeros of b^2-4b+4=16
The zeros of b^2-4b+4=16 are b=6,b=-2