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

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

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

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

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

x = 4, x = - 2

Solution steps

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

Expand

(x + 1) (x - 3) : 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} : 4

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

The solutions to the quadratic equation are:

x = 4, x = - 2

7 m^{2} - 8 mn + n^{2}

(5^{3})^{\frac{7}{6}}

5 y = 9 y^{2} - 4

(1 - i) (1 + i)

6 4^{\frac{1}{4}}

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