What are the solutions to the equation x^2-4x+4=81 ?

The solutions to the equation x^2-4x+4=81 are x=11,x=-7

Find the zeros of x^2-4x+4=81

The zeros of x^2-4x+4=81 are x=11,x=-7

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x^{2} - 4 x + 4 = 81

x = 11, x = - 7

Solution steps

x^{2} - 4 x + 4 = 81

Move

81

to the left side

x^{2} - 4 x - 77 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 1 \cdot ( - 77 )}{2 \cdot 1}

(- 4)^{2} - 4 \cdot 1 \cdot (- 77) = 18

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

Separate the solutions

x_{1} = \frac{- ( - 4 ) + 18}{2 \cdot 1}, x_{2} = \frac{- ( - 4 ) - 18}{2 \cdot 1}

x = \frac{- ( - 4 ) + 18}{2 \cdot 1} : 11

x = \frac{- ( - 4 ) - 18}{2 \cdot 1} : - 7

The solutions to the quadratic equation are:

x = 11, x = - 7

x^{2} - 15 x + 22 = 0

31 x^{2} + 62 x = 0

X^{2} - 6 X + 8 = 0

m^{2} - 4 m + 1 = 0

y, \frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1

What are the solutions to the equation x^2-4x+4=81 ?
The solutions to the equation x^2-4x+4=81 are x=11,x=-7
Find the zeros of x^2-4x+4=81
The zeros of x^2-4x+4=81 are x=11,x=-7