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

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

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

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

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

x = 15, x = 3

Solution steps

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

Expand

(x + 1)^{2} - 20 (x + 1) : x^{2} - 18 x - 19

x^{2} - 18 x - 19 = - 64

Move

64

to the left side

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

a c - a - b c + b + c^{2} - c

(x - 2)^{2} - 4 = x + 1

- \frac{2}{3} \cdot \frac{3}{2}

(x - 3)^{2} = 144

3^{\frac{1}{4} - 1}

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