What are the solutions to the equation x^2-18x+81=121 ?

The solutions to the equation x^2-18x+81=121 are x=20,x=-2

Find the zeros of x^2-18x+81=121

The zeros of x^2-18x+81=121 are x=20,x=-2

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x^{2} - 18 x + 81 = 121

x = 20, x = - 2

Solution steps

x^{2} - 18 x + 81 = 121

Move

121

to the left side

x^{2} - 18 x - 40 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 18 ) \pm ( - 18 ) ^{2} - 4 \cdot 1 \cdot ( - 40 )}{2 \cdot 1}

(- 18)^{2} - 4 \cdot 1 \cdot (- 40) = 22

x_{1, 2} = \frac{- ( - 18 ) \pm 22}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 18 ) + 22}{2 \cdot 1}, x_{2} = \frac{- ( - 18 ) - 22}{2 \cdot 1}

x = \frac{- ( - 18 ) + 22}{2 \cdot 1} : 20

x = \frac{- ( - 18 ) - 22}{2 \cdot 1} : - 2

The solutions to the quadratic equation are:

x = 20, x = - 2

x^{2} = 1 2^{2} + 5^{2}

11 x = x^{2} + 28

x^{2} - 20 x + 100 = 25

(3 x + 3)^{2} - 42 = - 39

4 x^{2} - 8 x

What are the solutions to the equation x^2-18x+81=121 ?
The solutions to the equation x^2-18x+81=121 are x=20,x=-2
Find the zeros of x^2-18x+81=121
The zeros of x^2-18x+81=121 are x=20,x=-2