What are the solutions to the equation 2x(x-4)=5 ?

The solutions to the equation 2x(x-4)=5 are x=(4+sqrt(26))/2 ,x=(4-sqrt(26))/2

Find the zeros of 2x(x-4)=5

The zeros of 2x(x-4)=5 are x=(4+sqrt(26))/2 ,x=(4-sqrt(26))/2

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2 x (x - 4) = 5

x = \frac{4 + 26}{2}, x = \frac{4 - 26}{2}

Decimal Notation

x = 4.54950 \dots, x = - 0.54950 \dots

Solution steps

2 x (x - 4) = 5

Expand

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

2 x^{2} - 8 x = 5

Move

5

to the left side

2 x^{2} - 8 x - 5 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 \cdot 2 ( - 5 )}{2 \cdot 2}

(- 8)^{2} - 4 \cdot 2 (- 5) = 226

x_{1, 2} = \frac{- ( - 8 ) \pm 2 26}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 8 ) + 2 26}{2 \cdot 2}, x_{2} = \frac{- ( - 8 ) - 2 26}{2 \cdot 2}

x = \frac{- ( - 8 ) + 2 26}{2 \cdot 2} : \frac{4 + 26}{2}

x = \frac{- ( - 8 ) - 2 26}{2 \cdot 2} : \frac{4 - 26}{2}

The solutions to the quadratic equation are:

x = \frac{4 + 26}{2}, x = \frac{4 - 26}{2}

x^{2} = 5^{2} + 1 0^{2}

12 = 4 x^{2} - 6 x + 8

x^{2} - 7 x - 3 = 0

x^{2} - 12 x + 51 = 2 x + 3 x

x^{2} - x = 2 x - 2

What are the solutions to the equation 2x(x-4)=5 ?
The solutions to the equation 2x(x-4)=5 are x=(4+sqrt(26))/2 ,x=(4-sqrt(26))/2
Find the zeros of 2x(x-4)=5
The zeros of 2x(x-4)=5 are x=(4+sqrt(26))/2 ,x=(4-sqrt(26))/2