What are the solutions to the equation 2x(x-5)=x+3 ?

The solutions to the equation 2x(x-5)=x+3 are x=(11+sqrt(145))/4 ,x=(11-sqrt(145))/4

Find the zeros of 2x(x-5)=x+3

The zeros of 2x(x-5)=x+3 are x=(11+sqrt(145))/4 ,x=(11-sqrt(145))/4

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2 x (x - 5) = x + 3

x = \frac{11 + 145}{4}, x = \frac{11 - 145}{4}

Decimal Notation

x = 5.76039 \dots, x = - 0.26039 \dots

Solution steps

2 x (x - 5) = x + 3

Expand

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

2 x^{2} - 10 x = x + 3

Move

3

to the left side

2 x^{2} - 10 x - 3 = x

Move

x

to the left side

2 x^{2} - 11 x - 3 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 11 ) \pm ( - 11 ) ^{2} - 4 \cdot 2 ( - 3 )}{2 \cdot 2}

(- 11)^{2} - 4 \cdot 2 (- 3) = 145

x_{1, 2} = \frac{- ( - 11 ) \pm 145}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 11 ) + 145}{2 \cdot 2}, x_{2} = \frac{- ( - 11 ) - 145}{2 \cdot 2}

x = \frac{- ( - 11 ) + 145}{2 \cdot 2} : \frac{11 + 145}{4}

x = \frac{- ( - 11 ) - 145}{2 \cdot 2} : \frac{11 - 145}{4}

The solutions to the quadratic equation are:

x = \frac{11 + 145}{4}, x = \frac{11 - 145}{4}

(a x + 7 b)^{2} - (b x + 7 a)^{2} = 0

- 9 v + 2 = v^{2} + 20

(x - 1)^{2} - 24 = 0

n^{2} + 5 n - 14 = 0

(x - 2) (3 - 2 x) = 3

What are the solutions to the equation 2x(x-5)=x+3 ?
The solutions to the equation 2x(x-5)=x+3 are x=(11+sqrt(145))/4 ,x=(11-sqrt(145))/4
Find the zeros of 2x(x-5)=x+3
The zeros of 2x(x-5)=x+3 are x=(11+sqrt(145))/4 ,x=(11-sqrt(145))/4