What are the solutions to the equation x^2-18x+81=49 ?

The solutions to the equation x^2-18x+81=49 are x=16,x=2

Find the zeros of x^2-18x+81=49

The zeros of x^2-18x+81=49 are x=16,x=2

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x^{2} - 18 x + 81 = 49

x = 16, x = 2

Solution steps

x^{2} - 18 x + 81 = 49

Move

49

to the left side

x^{2} - 18 x + 32 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 18 ) \pm ( - 18 ) ^{2} - 4 \cdot 1 \cdot 32}{2 \cdot 1}

(- 18)^{2} - 4 \cdot 1 \cdot 32 = 14

x_{1, 2} = \frac{- ( - 18 ) \pm 14}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 18 ) + 14}{2 \cdot 1}, x_{2} = \frac{- ( - 18 ) - 14}{2 \cdot 1}

x = \frac{- ( - 18 ) + 14}{2 \cdot 1} : 16

x = \frac{- ( - 18 ) - 14}{2 \cdot 1} : 2

The solutions to the quadratic equation are:

x = 16, x = 2

2 (x + 10)^{2} - 37 = 13

x^{2} - 72 = 0

x^{2} + x + 2

u, v^{2} = u^{2} + 2 a s

x^{2} - 16 = 0

What are the solutions to the equation x^2-18x+81=49 ?
The solutions to the equation x^2-18x+81=49 are x=16,x=2
Find the zeros of x^2-18x+81=49
The zeros of x^2-18x+81=49 are x=16,x=2