What are the solutions to the equation m^2-18m+81=100 ?

The solutions to the equation m^2-18m+81=100 are m=19,m=-1

Find the zeros of m^2-18m+81=100

The zeros of m^2-18m+81=100 are m=19,m=-1

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m^{2} - 18 m + 81 = 100

m = 19, m = - 1

Solution steps

m^{2} - 18 m + 81 = 100

Move

100

to the left side

m^{2} - 18 m - 19 = 0

Solve with the quadratic formula

m_{1, 2} = \frac{- ( - 18 ) \pm ( - 18 ) ^{2} - 4 \cdot 1 \cdot ( - 19 )}{2 \cdot 1}

(- 18)^{2} - 4 \cdot 1 \cdot (- 19) = 20

m_{1, 2} = \frac{- ( - 18 ) \pm 20}{2 \cdot 1}

Separate the solutions

m_{1} = \frac{- ( - 18 ) + 20}{2 \cdot 1}, m_{2} = \frac{- ( - 18 ) - 20}{2 \cdot 1}

m = \frac{- ( - 18 ) + 20}{2 \cdot 1} : 19

m = \frac{- ( - 18 ) - 20}{2 \cdot 1} : - 1

The solutions to the quadratic equation are:

m = 19, m = - 1

- 9 \leq \frac{1 - 2 x}{9} < 16

2 7^{\frac{1}{4}} \cdot 3^{\frac{1}{4}}

2 x^{2} + 23 = 14 x

\frac{5}{6} x = 3

z^{3} + 12 z^{2} + 27 z - 40

What are the solutions to the equation m^2-18m+81=100 ?
The solutions to the equation m^2-18m+81=100 are m=19,m=-1
Find the zeros of m^2-18m+81=100
The zeros of m^2-18m+81=100 are m=19,m=-1