What are the solutions to the equation (x+1)(x-4)=14 ?

The solutions to the equation (x+1)(x-4)=14 are x=6,x=-3

Find the zeros of (x+1)(x-4)=14

The zeros of (x+1)(x-4)=14 are x=6,x=-3

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(x + 1) (x - 4) = 14

x = 6, x = - 3

Solution steps

(x + 1) (x - 4) = 14

Expand

(x + 1) (x - 4) : x^{2} - 3 x - 4

x^{2} - 3 x - 4 = 14

Move

14

to the left side

x^{2} - 3 x - 18 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 1 \cdot ( - 18 )}{2 \cdot 1}

(- 3)^{2} - 4 \cdot 1 \cdot (- 18) = 9

x_{1, 2} = \frac{- ( - 3 ) \pm 9}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 3 ) + 9}{2 \cdot 1}, x_{2} = \frac{- ( - 3 ) - 9}{2 \cdot 1}

x = \frac{- ( - 3 ) + 9}{2 \cdot 1} : 6

x = \frac{- ( - 3 ) - 9}{2 \cdot 1} : - 3

The solutions to the quadratic equation are:

x = 6, x = - 3

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

- x^{2} + x = 1

3 x^{2} = - 2 x

2 x^{2} - 12 x + 17 = 0

3 x^{2} = - 3 x

What are the solutions to the equation (x+1)(x-4)=14 ?
The solutions to the equation (x+1)(x-4)=14 are x=6,x=-3
Find the zeros of (x+1)(x-4)=14
The zeros of (x+1)(x-4)=14 are x=6,x=-3