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

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

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

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

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

x = 5, x = - 3

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} : 5

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

The solutions to the quadratic equation are:

x = 5, x = - 3

5 x^{2} - 4 x - 1 = 0

4 x (x - 3) = 0

x, x^{2} - 9 y^{2} + 4 x + 4 = 0

x^{2} + x - 2 = 0

98 - 6 x^{2} = 0

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