What are the solutions to the equation 2x(x-5)=5x+1 ?

The solutions to the equation 2x(x-5)=5x+1 are x=(15+sqrt(233))/4 ,x=(15-sqrt(233))/4

Find the zeros of 2x(x-5)=5x+1

The zeros of 2x(x-5)=5x+1 are x=(15+sqrt(233))/4 ,x=(15-sqrt(233))/4

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

x = \frac{15 + 233}{4}, x = \frac{15 - 233}{4}

Decimal Notation

x = 7.56608 \dots, x = - 0.06608 \dots

Solution steps

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

Expand

2 x (x - 5) : 2 x^{2} - 10 x

2 x^{2} - 10 x = 5 x + 1

Move

1

to the left side

2 x^{2} - 10 x - 1 = 5 x

Move

5 x

to the left side

2 x^{2} - 15 x - 1 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 15 ) \pm ( - 15 ) ^{2} - 4 \cdot 2 ( - 1 )}{2 \cdot 2}

(- 15)^{2} - 4 \cdot 2 (- 1) = 233

x_{1, 2} = \frac{- ( - 15 ) \pm 233}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 15 ) + 233}{2 \cdot 2}, x_{2} = \frac{- ( - 15 ) - 233}{2 \cdot 2}

x = \frac{- ( - 15 ) + 233}{2 \cdot 2} : \frac{15 + 233}{4}

x = \frac{- ( - 15 ) - 233}{2 \cdot 2} : \frac{15 - 233}{4}

The solutions to the quadratic equation are:

x = \frac{15 + 233}{4}, x = \frac{15 - 233}{4}

m^{2} - 4 m + 4 = 8

x^{2} + 5 x - 225 = 0

(1 - 3 (- 1)) x^{2} + 3 x - 4 = 0

x, 6 x^{2} - 9 x + 1 = 0

- 4 x^{2} + x + 3 = 0

What are the solutions to the equation 2x(x-5)=5x+1 ?
The solutions to the equation 2x(x-5)=5x+1 are x=(15+sqrt(233))/4 ,x=(15-sqrt(233))/4
Find the zeros of 2x(x-5)=5x+1
The zeros of 2x(x-5)=5x+1 are x=(15+sqrt(233))/4 ,x=(15-sqrt(233))/4