What are the solutions to the equation (x-3)(x+1)=5 ?

The solutions to the equation (x-3)(x+1)=5 are x=4,x=-2

Find the zeros of (x-3)(x+1)=5

The zeros of (x-3)(x+1)=5 are x=4,x=-2

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(x - 3) (x + 1) = 5

x = 4, x = - 2

Solution steps

(x - 3) (x + 1) = 5

Expand

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

x^{2} - 2 x - 3 = 5

Move

5

to the left side

x^{2} - 2 x - 8 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 1 \cdot ( - 8 )}{2 \cdot 1}

(- 2)^{2} - 4 \cdot 1 \cdot (- 8) = 6

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

Separate the solutions

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

x = \frac{- ( - 2 ) + 6}{2 \cdot 1} : 4

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

The solutions to the quadratic equation are:

x = 4, x = - 2

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

2 (x + 1)^{2} = 50

λ^{2} - 6 λ + 9 (1) = 0

- 7 x^{2} + 6 x + 3 = 0

6 x = 27 - x^{2}

What are the solutions to the equation (x-3)(x+1)=5 ?
The solutions to the equation (x-3)(x+1)=5 are x=4,x=-2
Find the zeros of (x-3)(x+1)=5
The zeros of (x-3)(x+1)=5 are x=4,x=-2