What are the solutions to the equation (x(x-8))/4+2=0 ?

The solutions to the equation (x(x-8))/4+2=0 are x=2(2+sqrt(2)),x=2(2-sqrt(2))

Find the zeros of (x(x-8))/4+2=0

The zeros of (x(x-8))/4+2=0 are x=2(2+sqrt(2)),x=2(2-sqrt(2))

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\frac{x ( x - 8 )}{4} + 2 = 0

x = 2 (2 + 2), x = 2 (2 - 2)

Decimal Notation

x = 6.82842 \dots, x = 1.17157 \dots

Solution steps

\frac{x ( x - 8 )}{4} + 2 = 0

Multiply both sides by

4

x (x - 8) + 8 = 0

Expand

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

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

Solve with the quadratic formula

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

(- 8)^{2} - 4 \cdot 1 \cdot 8 = 42

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

Separate the solutions

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

x = \frac{- ( - 8 ) + 4 2}{2 \cdot 1} : 2 (2 + 2)

x = \frac{- ( - 8 ) - 4 2}{2 \cdot 1} : 2 (2 - 2)

The solutions to the quadratic equation are:

x = 2 (2 + 2), x = 2 (2 - 2)

x^{2} - 8 x + 16 = 8

10 x^{2} - 10 = 0

3 x^{2} + 5 x - 23 = 0

x^{2} - 5 x - 6 = 0

8 y^{2} + 128 y - 512 = 0

What are the solutions to the equation (x(x-8))/4+2=0 ?
The solutions to the equation (x(x-8))/4+2=0 are x=2(2+sqrt(2)),x=2(2-sqrt(2))
Find the zeros of (x(x-8))/4+2=0
The zeros of (x(x-8))/4+2=0 are x=2(2+sqrt(2)),x=2(2-sqrt(2))