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

The solutions to the equation x^2-(3x+1)/2 = 2/3 are x=(9+sqrt(249))/(12),x=(9-sqrt(249))/(12)

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

The zeros of x^2-(3x+1)/2 = 2/3 are x=(9+sqrt(249))/(12),x=(9-sqrt(249))/(12)

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x^{2} - \frac{3 x + 1}{2} = \frac{2}{3}

x = \frac{9 + 249}{12}, x = \frac{9 - 249}{12}

Decimal Notation

x = 2.06497 \dots, x = - 0.56497 \dots

Solution steps

x^{2} - \frac{3 x + 1}{2} = \frac{2}{3}

Find Least Common Multiplier of

2, 3 : 6

Multiply by LCM=

6

x^{2} \cdot 6 - \frac{3 x + 1}{2} \cdot 6 = \frac{2}{3} \cdot 6

Simplify

6 x^{2} - 3 (3 x + 1) = 4

Expand

6 x^{2} - 3 (3 x + 1) : 6 x^{2} - 9 x - 3

6 x^{2} - 9 x - 3 = 4

Move

4

to the left side

6 x^{2} - 9 x - 7 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 9 ) \pm ( - 9 ) ^{2} - 4 \cdot 6 ( - 7 )}{2 \cdot 6}

(- 9)^{2} - 4 \cdot 6 (- 7) = 249

x_{1, 2} = \frac{- ( - 9 ) \pm 249}{2 \cdot 6}

Separate the solutions

x_{1} = \frac{- ( - 9 ) + 249}{2 \cdot 6}, x_{2} = \frac{- ( - 9 ) - 249}{2 \cdot 6}

x = \frac{- ( - 9 ) + 249}{2 \cdot 6} : \frac{9 + 249}{12}

x = \frac{- ( - 9 ) - 249}{2 \cdot 6} : \frac{9 - 249}{12}

The solutions to the quadratic equation are:

x = \frac{9 + 249}{12}, x = \frac{9 - 249}{12}

y^{2} - y - 24 = 0

0 = x^{2} + 2 x + 5

(x + 2)^{2} = a

- x^{2} + 6 x = x

2 x^{2} - 10 = - x

What are the solutions to the equation x^2-(3x+1)/2 = 2/3 ?
The solutions to the equation x^2-(3x+1)/2 = 2/3 are x=(9+sqrt(249))/(12),x=(9-sqrt(249))/(12)
Find the zeros of x^2-(3x+1)/2 = 2/3
The zeros of x^2-(3x+1)/2 = 2/3 are x=(9+sqrt(249))/(12),x=(9-sqrt(249))/(12)