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

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

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

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

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

x = 3, x = - 6

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} : 3

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

The solutions to the quadratic equation are:

x = 3, x = - 6

2 (- 1)^{3} - 12 (- 1)^{2} - 30 (- 1) + 9

\frac{( 2 n ) !}{n ! n !}

43 + 211

(\frac{1}{1.1})^{2}

x^{24}

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