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

The solutions to the equation 2x(x-3)=1 are x=(3+sqrt(11))/2 ,x=(3-sqrt(11))/2

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

The zeros of 2x(x-3)=1 are x=(3+sqrt(11))/2 ,x=(3-sqrt(11))/2

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2 x (x - 3) = 1

x = \frac{3 + 11}{2}, x = \frac{3 - 11}{2}

Solution steps

2 x (x - 3) = 1

Expand

2 x (x - 3) : 2 x^{2} - 6 x

2 x^{2} - 6 x = 1

Move

1

to the left side

2 x^{2} - 6 x - 1 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 6 ) \pm ( - 6 ) ^{2} - 4 \cdot 2 ( - 1 )}{2 \cdot 2}

(- 6)^{2} - 4 \cdot 2 (- 1) = 211

x_{1, 2} = \frac{- ( - 6 ) \pm 2 11}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 6 ) + 2 11}{2 \cdot 2}, x_{2} = \frac{- ( - 6 ) - 2 11}{2 \cdot 2}

x = \frac{- ( - 6 ) + 2 11}{2 \cdot 2} : \frac{3 + 11}{2}

x = \frac{- ( - 6 ) - 2 11}{2 \cdot 2} : \frac{3 - 11}{2}

The solutions to the quadratic equation are:

x = \frac{3 + 11}{2}, x = \frac{3 - 11}{2}

5 (p^{2} - 4) - 1 = 4

- 4 x^{2} + 8 x = 7

3 - 2 x^{2} = - 5

\frac{7}{5} x^{2} = 2

y^{2} - 12 y = - 25

What are the solutions to the equation 2x(x-3)=1 ?
The solutions to the equation 2x(x-3)=1 are x=(3+sqrt(11))/2 ,x=(3-sqrt(11))/2
Find the zeros of 2x(x-3)=1
The zeros of 2x(x-3)=1 are x=(3+sqrt(11))/2 ,x=(3-sqrt(11))/2