What are the solutions to the equation r^2-10r+25=1 ?

The solutions to the equation r^2-10r+25=1 are r=6,r=4

Find the zeros of r^2-10r+25=1

The zeros of r^2-10r+25=1 are r=6,r=4

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r^{2} - 10 r + 25 = 1

r = 6, r = 4

Solution steps

r^{2} - 10 r + 25 = 1

Move

1

to the left side

r^{2} - 10 r + 24 = 0

Solve with the quadratic formula

r_{1, 2} = \frac{- ( - 10 ) \pm ( - 10 ) ^{2} - 4 \cdot 1 \cdot 24}{2 \cdot 1}

(- 10)^{2} - 4 \cdot 1 \cdot 24 = 2

r_{1, 2} = \frac{- ( - 10 ) \pm 2}{2 \cdot 1}

Separate the solutions

r_{1} = \frac{- ( - 10 ) + 2}{2 \cdot 1}, r_{2} = \frac{- ( - 10 ) - 2}{2 \cdot 1}

r = \frac{- ( - 10 ) + 2}{2 \cdot 1} : 6

r = \frac{- ( - 10 ) - 2}{2 \cdot 1} : 4

The solutions to the quadratic equation are:

r = 6, r = 4

(6 x - 7) = (4 x + 23)

(- \frac{4}{5})^{0}

\frac{4}{x + 3} \geq 1

(1 + \frac{0.06}{4})^{4} - 1

(w + 4)^{2} + 95 = 5

What are the solutions to the equation r^2-10r+25=1 ?
The solutions to the equation r^2-10r+25=1 are r=6,r=4
Find the zeros of r^2-10r+25=1
The zeros of r^2-10r+25=1 are r=6,r=4