What are the solutions to the equation (x-2)(x+1)=4 ?

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

Find the zeros of (x-2)(x+1)=4

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

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(x - 2) (x + 1) = 4

x = 3, x = - 2

Solution steps

(x - 2) (x + 1) = 4

Expand

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

x^{2} - x - 2 = 4

Move

4

to the left side

x^{2} - x - 6 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 1 \cdot ( - 6 )}{2 \cdot 1}

(- 1)^{2} - 4 \cdot 1 \cdot (- 6) = 5

x_{1, 2} = \frac{- ( - 1 ) \pm 5}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 1 ) + 5}{2 \cdot 1}, x_{2} = \frac{- ( - 1 ) - 5}{2 \cdot 1}

x = \frac{- ( - 1 ) + 5}{2 \cdot 1} : 3

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

The solutions to the quadratic equation are:

x = 3, x = - 2

3 x^{2} - 4 x = - 8

- \frac{7}{8} + \frac{1}{4}

4 x - 2 x^{2} \geq 0

25 + 36

3.5 n - 1356 - 1.28 \cdot 2.916 n > 0

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