What are the solutions to the equation (x+3)(x-1)=5 ?

The solutions to the equation (x+3)(x-1)=5 are x=2,x=-4

Find the zeros of (x+3)(x-1)=5

The zeros of (x+3)(x-1)=5 are x=2,x=-4

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(x + 3) (x - 1) = 5

x = 2, x = - 4

Solution steps

(x + 3) (x - 1) = 5

Expand

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

x^{2} + 2 x - 3 = 5

Move

5

to the left side

x^{2} + 2 x - 8 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 1 \cdot ( - 8 )}{2 \cdot 1}

2^{2} - 4 \cdot 1 \cdot (- 8) = 6

x_{1, 2} = \frac{- 2 \pm 6}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 2 + 6}{2 \cdot 1}, x_{2} = \frac{- 2 - 6}{2 \cdot 1}

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

x = \frac{- 2 - 6}{2 \cdot 1} : - 4

The solutions to the quadratic equation are:

x = 2, x = - 4

(1 + \frac{2 i}{n})^{2}

lo g_{a} (18) \cdot lo g_{3} (a)

(x - 2) (x + 5) < 0

- 3 a + 10 a = - 35

- 3 x^{2} - 18 x - 35 = 0

What are the solutions to the equation (x+3)(x-1)=5 ?
The solutions to the equation (x+3)(x-1)=5 are x=2,x=-4
Find the zeros of (x+3)(x-1)=5
The zeros of (x+3)(x-1)=5 are x=2,x=-4