What are the solutions to the equation 121+x^2=22x ?

The solutions to the equation 121+x^2=22x are x=11

Find the zeros of 121+x^2=22x

The zeros of 121+x^2=22x are x=11

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121 + x^{2} = 22 x

x = 11

Solution steps

121 + x^{2} = 22 x

Move

22 x

to the left side

121 + x^{2} - 22 x = 0

Write in the standard form

a x^{2} + b x + c = 0

x^{2} - 22 x + 121 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 22 ) \pm ( - 22 ) ^{2} - 4 \cdot 1 \cdot 121}{2 \cdot 1}

(- 22)^{2} - 4 \cdot 1 \cdot 121 = 0

x_{1, 2} = \frac{- ( - 22 ) \pm 0}{2 \cdot 1}

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

\frac{- ( - 22 )}{2 \cdot 1} = 11

x = 11

The solution to the quadratic equation is:

x = 11

- 8 (v + 1) = 2 (1 - 4 v) - 9

x^{2} + 1 = 2 x

5 r^{2} - 45 = 0

x^{2} - 36 > 0

- 4 (x + 3) = 16 - 4 x

What are the solutions to the equation 121+x^2=22x ?
The solutions to the equation 121+x^2=22x are x=11
Find the zeros of 121+x^2=22x
The zeros of 121+x^2=22x are x=11