What are the solutions to the equation (x+1)^2-20(x+1)+64=0 ?

The solutions to the equation (x+1)^2-20(x+1)+64=0 are x=15,x=3

Find the zeros of (x+1)^2-20(x+1)+64=0

The zeros of (x+1)^2-20(x+1)+64=0 are x=15,x=3

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(x + 1)^{2} - 20 (x + 1) + 64 = 0

x = 15, x = 3

Solution steps

(x + 1)^{2} - 20 (x + 1) + 64 = 0

Expand

(x + 1)^{2} - 20 (x + 1) + 64 : x^{2} - 18 x + 45

x^{2} - 18 x + 45 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 18 ) \pm ( - 18 ) ^{2} - 4 \cdot 1 \cdot 45}{2 \cdot 1}

(- 18)^{2} - 4 \cdot 1 \cdot 45 = 12

x_{1, 2} = \frac{- ( - 18 ) \pm 12}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 18 ) + 12}{2 \cdot 1}, x_{2} = \frac{- ( - 18 ) - 12}{2 \cdot 1}

x = \frac{- ( - 18 ) + 12}{2 \cdot 1} : 15

x = \frac{- ( - 18 ) - 12}{2 \cdot 1} : 3

The solutions to the quadratic equation are:

x = 15, x = 3

(5 + 2 i) (3 - 4 i)

- 3 + 3 x = - 2 (x + 1)

9 x - 17 + 3 x + 19 + 2 x + 10 = 180

0 = x^{2} + 6 x + 4

- \frac{7}{3} - \frac{7}{3}

What are the solutions to the equation (x+1)^2-20(x+1)+64=0 ?
The solutions to the equation (x+1)^2-20(x+1)+64=0 are x=15,x=3
Find the zeros of (x+1)^2-20(x+1)+64=0
The zeros of (x+1)^2-20(x+1)+64=0 are x=15,x=3