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

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

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

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

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x^{2} + 121 = 22 x

x = 11

Solution steps

x^{2} + 121 = 22 x

Move

22 x

to the left side

x^{2} + 121 - 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

v (3 v - 5) = - 2

y^{2} + 4 y - 12 = 0

a, a^{2} + b^{2} = 0

(x - 9)^{2} = 169

y, z = y^{2}

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