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

The solutions to the equation 3x(2x-5)=3 are x=(5+sqrt(33))/4 ,x=(5-sqrt(33))/4

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

The zeros of 3x(2x-5)=3 are x=(5+sqrt(33))/4 ,x=(5-sqrt(33))/4

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3 x (2 x - 5) = 3

x = \frac{5 + 33}{4}, x = \frac{5 - 33}{4}

Solution steps

3 x (2 x - 5) = 3

Expand

3 x (2 x - 5) : 6 x^{2} - 15 x

6 x^{2} - 15 x = 3

Move

3

to the left side

6 x^{2} - 15 x - 3 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 15 ) \pm ( - 15 ) ^{2} - 4 \cdot 6 ( - 3 )}{2 \cdot 6}

(- 15)^{2} - 4 \cdot 6 (- 3) = 333

x_{1, 2} = \frac{- ( - 15 ) \pm 3 33}{2 \cdot 6}

Separate the solutions

x_{1} = \frac{- ( - 15 ) + 3 33}{2 \cdot 6}, x_{2} = \frac{- ( - 15 ) - 3 33}{2 \cdot 6}

x = \frac{- ( - 15 ) + 3 33}{2 \cdot 6} : \frac{5 + 33}{4}

x = \frac{- ( - 15 ) - 3 33}{2 \cdot 6} : \frac{5 - 33}{4}

The solutions to the quadratic equation are:

x = \frac{5 + 33}{4}, x = \frac{5 - 33}{4}

2 s^{2} - 98 = 0

6 x^{2} + 23 x - 4 = 0

67 = \frac{v ^{2}}{5.7}

(2 x + 1) (x - 1) = (3 x - 1) (x + 1)

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

What are the solutions to the equation 3x(2x-5)=3 ?
The solutions to the equation 3x(2x-5)=3 are x=(5+sqrt(33))/4 ,x=(5-sqrt(33))/4
Find the zeros of 3x(2x-5)=3
The zeros of 3x(2x-5)=3 are x=(5+sqrt(33))/4 ,x=(5-sqrt(33))/4