What are the solutions to the equation (x-1)(x+3)=12 ?

The solutions to the equation (x-1)(x+3)=12 are x=3,x=-5

Find the zeros of (x-1)(x+3)=12

The zeros of (x-1)(x+3)=12 are x=3,x=-5

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(x - 1) (x + 3) = 12

x = 3, x = - 5

Solution steps

(x - 1) (x + 3) = 12

Expand

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

x^{2} + 2 x - 3 = 12

Move

12

to the left side

x^{2} + 2 x - 15 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 1 \cdot ( - 15 )}{2 \cdot 1}

2^{2} - 4 \cdot 1 \cdot (- 15) = 8

x_{1, 2} = \frac{- 2 \pm 8}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 2 + 8}{2 \cdot 1}, x_{2} = \frac{- 2 - 8}{2 \cdot 1}

x = \frac{- 2 + 8}{2 \cdot 1} : 3

x = \frac{- 2 - 8}{2 \cdot 1} : - 5

The solutions to the quadratic equation are:

x = 3, x = - 5

x^{2} - 4 x + 3

x^{2} - 8 x + 4 > - 3

\frac{x + 1}{x} > 2

9 x^{2} + 6 x + 9

2 k^{2} + 9 k + 10

What are the solutions to the equation (x-1)(x+3)=12 ?
The solutions to the equation (x-1)(x+3)=12 are x=3,x=-5
Find the zeros of (x-1)(x+3)=12
The zeros of (x-1)(x+3)=12 are x=3,x=-5