What are the solutions to the equation x^2-14x+49=1 ?

The solutions to the equation x^2-14x+49=1 are x=8,x=6

Find the zeros of x^2-14x+49=1

The zeros of x^2-14x+49=1 are x=8,x=6

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x^{2} - 14 x + 49 = 1

x = 8, x = 6

Solution steps

x^{2} - 14 x + 49 = 1

Move

1

to the left side

x^{2} - 14 x + 48 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 14 ) \pm ( - 14 ) ^{2} - 4 \cdot 1 \cdot 48}{2 \cdot 1}

(- 14)^{2} - 4 \cdot 1 \cdot 48 = 2

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

Separate the solutions

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

x = \frac{- ( - 14 ) + 2}{2 \cdot 1} : 8

x = \frac{- ( - 14 ) - 2}{2 \cdot 1} : 6

The solutions to the quadratic equation are:

x = 8, x = 6

1 = - \frac{1}{2} + \frac{x}{e ^{- \frac{3}{10}}}

(\frac{1}{t ^{- 7} \cdot t ^{- 1}})^{\frac{5}{16}}

z = \frac{1 - i}{1 + i}

\frac{∣ x - 2 ∣}{4} \geq 5

5 y^{2} + 3 y - 2

What are the solutions to the equation x^2-14x+49=1 ?
The solutions to the equation x^2-14x+49=1 are x=8,x=6
Find the zeros of x^2-14x+49=1
The zeros of x^2-14x+49=1 are x=8,x=6