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

The solutions to the equation (x+3)(x-5)=1 are x=1+sqrt(17),x=1-sqrt(17)

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

The zeros of (x+3)(x-5)=1 are x=1+sqrt(17),x=1-sqrt(17)

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

x = 1 + 17, x = 1 - 17

Decimal Notation

x = 5.12310 \dots, x = - 3.12310 \dots

Solution steps

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

Expand

(x + 3) (x - 5) : x^{2} - 2 x - 15

x^{2} - 2 x - 15 = 1

Move

1

to the left side

x^{2} - 2 x - 16 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 1 \cdot ( - 16 )}{2 \cdot 1}

(- 2)^{2} - 4 \cdot 1 \cdot (- 16) = 217

x_{1, 2} = \frac{- ( - 2 ) \pm 2 17}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 2 ) + 2 17}{2 \cdot 1}, x_{2} = \frac{- ( - 2 ) - 2 17}{2 \cdot 1}

x = \frac{- ( - 2 ) + 2 17}{2 \cdot 1} : 1 + 17

x = \frac{- ( - 2 ) - 2 17}{2 \cdot 1} : 1 - 17

The solutions to the quadratic equation are:

x = 1 + 17, x = 1 - 17

2 (3 k - 1)^{2} = 18

- 3 x^{2} + 6 x = - 1

(- 2)^{3} = (- 2) x (- 2) x (- 2)

4 s^{2} + 4 s + 5 = 0

0 = - 2.7 t^{2} + 30 t + 6.5

What are the solutions to the equation (x+3)(x-5)=1 ?
The solutions to the equation (x+3)(x-5)=1 are x=1+sqrt(17),x=1-sqrt(17)
Find the zeros of (x+3)(x-5)=1
The zeros of (x+3)(x-5)=1 are x=1+sqrt(17),x=1-sqrt(17)