What are the solutions to the equation 5x(x-3)-(x-1)^2=2x ?

The solutions to the equation 5x(x-3)-(x-1)^2=2x are x=(15+sqrt(241))/8 ,x=(15-sqrt(241))/8

Find the zeros of 5x(x-3)-(x-1)^2=2x

The zeros of 5x(x-3)-(x-1)^2=2x are x=(15+sqrt(241))/8 ,x=(15-sqrt(241))/8

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5 x (x - 3) - (x - 1)^{2} = 2 x

x = \frac{15 + 241}{8}, x = \frac{15 - 241}{8}

Decimal Notation

x = 3.81552 \dots, x = - 0.06552 \dots

Solution steps

5 x (x - 3) - (x - 1)^{2} = 2 x

Expand

5 x (x - 3) - (x - 1)^{2} : 4 x^{2} - 13 x - 1

4 x^{2} - 13 x - 1 = 2 x

Move

2 x

to the left side

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

Solve with the quadratic formula

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

(- 15)^{2} - 4 \cdot 4 (- 1) = 241

x_{1, 2} = \frac{- ( - 15 ) \pm 241}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- ( - 15 ) + 241}{2 \cdot 4}, x_{2} = \frac{- ( - 15 ) - 241}{2 \cdot 4}

x = \frac{- ( - 15 ) + 241}{2 \cdot 4} : \frac{15 + 241}{8}

x = \frac{- ( - 15 ) - 241}{2 \cdot 4} : \frac{15 - 241}{8}

The solutions to the quadratic equation are:

x = \frac{15 + 241}{8}, x = \frac{15 - 241}{8}

- 25 x^{2} - 20 x - 4 = 0

x^{2} - 2 x - 15

lo g_{2} (x) + lo g_{2} (x - 1) < 1

125 e^{0.7}

\frac{8}{( 2 + 2 ) ^{3}}

What are the solutions to the equation 5x(x-3)-(x-1)^2=2x ?
The solutions to the equation 5x(x-3)-(x-1)^2=2x are x=(15+sqrt(241))/8 ,x=(15-sqrt(241))/8
Find the zeros of 5x(x-3)-(x-1)^2=2x
The zeros of 5x(x-3)-(x-1)^2=2x are x=(15+sqrt(241))/8 ,x=(15-sqrt(241))/8