What are the solutions to the equation (x+3)(x-5)=-8 ?

The solutions to the equation (x+3)(x-5)=-8 are x=1+2sqrt(2),x=1-2sqrt(2)

Find the zeros of (x+3)(x-5)=-8

The zeros of (x+3)(x-5)=-8 are x=1+2sqrt(2),x=1-2sqrt(2)

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(x + 3) (x - 5) = - 8

x = 1 + 22, x = 1 - 22

Decimal Notation

x = 3.82842 \dots, x = - 1.82842 \dots

Solution steps

(x + 3) (x - 5) = - 8

Expand

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

x^{2} - 2 x - 15 = - 8

Move

8

to the left side

x^{2} - 2 x - 7 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 1 \cdot ( - 7 )}{2 \cdot 1}

(- 2)^{2} - 4 \cdot 1 \cdot (- 7) = 42

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

Separate the solutions

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

x = \frac{- ( - 2 ) + 4 2}{2 \cdot 1} : 1 + 22

x = \frac{- ( - 2 ) - 4 2}{2 \cdot 1} : 1 - 22

The solutions to the quadratic equation are:

x = 1 + 22, x = 1 - 22

x^{2} - \frac{3}{2} = 0

x, (5 p - 52) x^{2} - (p - 4) x + 4 = 0

4 x^{2} - 15 x + 10 = 0

x^{2} - 226 x + 1105 = 0

x, x^{2} y - 4 x + 4 = 0

What are the solutions to the equation (x+3)(x-5)=-8 ?
The solutions to the equation (x+3)(x-5)=-8 are x=1+2sqrt(2),x=1-2sqrt(2)
Find the zeros of (x+3)(x-5)=-8
The zeros of (x+3)(x-5)=-8 are x=1+2sqrt(2),x=1-2sqrt(2)