What are the solutions to the equation r^2-8r+16=0 ?

The solutions to the equation r^2-8r+16=0 are r=4

Find the zeros of r^2-8r+16=0

The zeros of r^2-8r+16=0 are r=4

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r^{2} - 8 r + 16 = 0

r = 4

Solution steps

r^{2} - 8 r + 16 = 0

Solve with the quadratic formula

r_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 \cdot 1 \cdot 16}{2 \cdot 1}

(- 8)^{2} - 4 \cdot 1 \cdot 16 = 0

r_{1, 2} = \frac{- ( - 8 ) \pm 0}{2 \cdot 1}

r = \frac{- ( - 8 )}{2 \cdot 1}

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

r = 4

The solution to the quadratic equation is:

r = 4

∣ - 7 ∣ + 3

x + 3.8 > - 10.02

(\frac{- 1}{3})^{2}

n^{3} - n^{2} - 42 n

9 x^{2} - 6 x + 226 = 0

What are the solutions to the equation r^2-8r+16=0 ?
The solutions to the equation r^2-8r+16=0 are r=4
Find the zeros of r^2-8r+16=0
The zeros of r^2-8r+16=0 are r=4