What are the solutions to the equation (4-r)(-4-r)+16=0 ?

The solutions to the equation (4-r)(-4-r)+16=0 are r=0

Find the zeros of (4-r)(-4-r)+16=0

The zeros of (4-r)(-4-r)+16=0 are r=0

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(4 - r) (- 4 - r) + 16 = 0

r = 0

Solution steps

(4 - r) (- 4 - r) + 16 = 0

Expand

(4 - r) (- 4 - r) + 16 : r^{2}

r^{2} = 0

Solve with the quadratic formula

r_{1, 2} = \frac{- 0 \pm 0 ^{2} - 4 \cdot 1 \cdot 0}{2 \cdot 1}

0^{2} - 4 \cdot 1 \cdot 0 = 0

r_{1, 2} = \frac{- 0 \pm 0}{2 \cdot 1}

r = \frac{- 0}{2 \cdot 1}

\frac{- 0}{2 \cdot 1} = 0

r = 0

The solution to the quadratic equation is:

r = 0

c^{2} - 5 c + 3 = 0

(z^{2} + 1) = 0

3 x^{2} + 18 x - 45 = 0

x^{2} + 18 x + 92 = 0

7 x^{2} + 21 x = 28

What are the solutions to the equation (4-r)(-4-r)+16=0 ?
The solutions to the equation (4-r)(-4-r)+16=0 are r=0
Find the zeros of (4-r)(-4-r)+16=0
The zeros of (4-r)(-4-r)+16=0 are r=0