What are the solutions to the equation (x+1)^2-8*(x+1)+7=0 ?

The solutions to the equation (x+1)^2-8*(x+1)+7=0 are x=6,x=0

Find the zeros of (x+1)^2-8*(x+1)+7=0

The zeros of (x+1)^2-8*(x+1)+7=0 are x=6,x=0

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(x + 1)^{2} - 8 \cdot (x + 1) + 7 = 0

x = 6, x = 0

Solution steps

(x + 1)^{2} - 8 (x + 1) + 7 = 0

Expand

(x + 1)^{2} - 8 (x + 1) + 7 : x^{2} - 6 x

x^{2} - 6 x = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 6 ) \pm ( - 6 ) ^{2} - 4 \cdot 1 \cdot 0}{2 \cdot 1}

(- 6)^{2} - 4 \cdot 1 \cdot 0 = 6

x_{1, 2} = \frac{- ( - 6 ) \pm 6}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 6 ) + 6}{2 \cdot 1}, x_{2} = \frac{- ( - 6 ) - 6}{2 \cdot 1}

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

x = \frac{- ( - 6 ) - 6}{2 \cdot 1} : 0

The solutions to the quadratic equation are:

x = 6, x = 0

x, x^{2} + (y - 32 x)^{2} = 1

x^{2} - x - 48 = 0

- 0.01 x^{2} - 0.2 x + 8 = 0

x, y = - 0.00001 x^{2} + 0.01 x + 0.2001

2 m^{2} - 13 m - 45 = 0

What are the solutions to the equation (x+1)^2-8*(x+1)+7=0 ?
The solutions to the equation (x+1)^2-8*(x+1)+7=0 are x=6,x=0
Find the zeros of (x+1)^2-8*(x+1)+7=0
The zeros of (x+1)^2-8*(x+1)+7=0 are x=6,x=0