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

The solutions to the equation (2x-1)(3x-2)=3 are x=(7+sqrt(73))/(12),x=(7-sqrt(73))/(12)

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

The zeros of (2x-1)(3x-2)=3 are x=(7+sqrt(73))/(12),x=(7-sqrt(73))/(12)

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

x = \frac{7 + 73}{12}, x = \frac{7 - 73}{12}

Decimal Notation

x = 1.29533 \dots, x = - 0.12866 \dots

Solution steps

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

Expand

(2 x - 1) (3 x - 2) : 6 x^{2} - 7 x + 2

6 x^{2} - 7 x + 2 = 3

Move

3

to the left side

6 x^{2} - 7 x - 1 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 \cdot 6 ( - 1 )}{2 \cdot 6}

(- 7)^{2} - 4 \cdot 6 (- 1) = 73

x_{1, 2} = \frac{- ( - 7 ) \pm 73}{2 \cdot 6}

Separate the solutions

x_{1} = \frac{- ( - 7 ) + 73}{2 \cdot 6}, x_{2} = \frac{- ( - 7 ) - 73}{2 \cdot 6}

x = \frac{- ( - 7 ) + 73}{2 \cdot 6} : \frac{7 + 73}{12}

x = \frac{- ( - 7 ) - 73}{2 \cdot 6} : \frac{7 - 73}{12}

The solutions to the quadratic equation are:

x = \frac{7 + 73}{12}, x = \frac{7 - 73}{12}

x, x^{2} + y^{2} + z^{2} + 9 x - 2 y + 10 z + 19 = 0

\frac{n ( n - 3 )}{2} = 4 n

3 x^{2} + 8 x - 14 = 0

\frac{n ( n - 3 )}{2} = 3 n

70 = 8 b^{2} - 2

What are the solutions to the equation (2x-1)(3x-2)=3 ?
The solutions to the equation (2x-1)(3x-2)=3 are x=(7+sqrt(73))/(12),x=(7-sqrt(73))/(12)
Find the zeros of (2x-1)(3x-2)=3
The zeros of (2x-1)(3x-2)=3 are x=(7+sqrt(73))/(12),x=(7-sqrt(73))/(12)