What are the solutions to the equation (x-2)(x+2)-7(x-1)=21 ?

The solutions to the equation (x-2)(x+2)-7(x-1)=21 are x=9,x=-2

Find the zeros of (x-2)(x+2)-7(x-1)=21

The zeros of (x-2)(x+2)-7(x-1)=21 are x=9,x=-2

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(x - 2) (x + 2) - 7 (x - 1) = 21

x = 9, x = - 2

Solution steps

(x - 2) (x + 2) - 7 (x - 1) = 21

Expand

(x - 2) (x + 2) - 7 (x - 1) : x^{2} - 7 x + 3

x^{2} - 7 x + 3 = 21

Move

21

to the left side

x^{2} - 7 x - 18 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 \cdot 1 \cdot ( - 18 )}{2 \cdot 1}

(- 7)^{2} - 4 \cdot 1 \cdot (- 18) = 11

x_{1, 2} = \frac{- ( - 7 ) \pm 11}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 7 ) + 11}{2 \cdot 1}, x_{2} = \frac{- ( - 7 ) - 11}{2 \cdot 1}

x = \frac{- ( - 7 ) + 11}{2 \cdot 1} : 9

x = \frac{- ( - 7 ) - 11}{2 \cdot 1} : - 2

The solutions to the quadratic equation are:

x = 9, x = - 2

5 - 3 \cdot \frac{z - z ^{- 1}}{2 \cdot i}

\frac{2}{1 + 2}

11 x + 11 = - 2 x - 8

b^{2} + 1 1^{2} = 2 1^{2}

- 10 + 3 x = 5

What are the solutions to the equation (x-2)(x+2)-7(x-1)=21 ?
The solutions to the equation (x-2)(x+2)-7(x-1)=21 are x=9,x=-2
Find the zeros of (x-2)(x+2)-7(x-1)=21
The zeros of (x-2)(x+2)-7(x-1)=21 are x=9,x=-2