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

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

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

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

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

x = 2, x = - 3

Solution steps

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

Expand

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

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

The solutions to the quadratic equation are:

x = 2, x = - 3

x^{2} + 8 = 89

x^{2} + 8 = 15

1 + 2 + 3 + n = \frac{1}{2} n (n + 1)

4 x^{2} + 20 x + 45 = 0

(x - 4)^{2} + 5 x (x - 1) = 16

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