What are the solutions to the equation (x-8)(x+4)=1 ?

The solutions to the equation (x-8)(x+4)=1 are x=2+sqrt(37),x=2-sqrt(37)

Find the zeros of (x-8)(x+4)=1

The zeros of (x-8)(x+4)=1 are x=2+sqrt(37),x=2-sqrt(37)

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(x - 8) (x + 4) = 1

x = 2 + 37, x = 2 - 37

Decimal

x = 8.08276 \dots, x = - 4.08276 \dots

Solution steps

(x - 8) (x + 4) = 1

Expand

(x - 8) (x + 4) : x^{2} - 4 x - 32

x^{2} - 4 x - 32 = 1

Move

1

to the left side

x^{2} - 4 x - 33 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 1 \cdot ( - 33 )}{2 \cdot 1}

(- 4)^{2} - 4 \cdot 1 \cdot (- 33) = 237

x_{1, 2} = \frac{- ( - 4 ) \pm 2 37}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 4 ) + 2 37}{2 \cdot 1}, x_{2} = \frac{- ( - 4 ) - 2 37}{2 \cdot 1}

x = \frac{- ( - 4 ) + 2 37}{2 \cdot 1} : 2 + 37

x = \frac{- ( - 4 ) - 2 37}{2 \cdot 1} : 2 - 37

The solutions to the quadratic equation are:

x = 2 + 37, x = 2 - 37

x^{2} - 15 x + 49 = 0

(0.3 - x) (1.5 - x) + 0.9 (0.375) = 0

3 x^{2} - 3 x - 216 = 0

- 3 y^{2} + 8 y - 3 = 0

x, f (x + 3) = x^{2}

What are the solutions to the equation (x-8)(x+4)=1 ?
The solutions to the equation (x-8)(x+4)=1 are x=2+sqrt(37),x=2-sqrt(37)
Find the zeros of (x-8)(x+4)=1
The zeros of (x-8)(x+4)=1 are x=2+sqrt(37),x=2-sqrt(37)