What are the solutions to the equation 3x^2+9/2 x-4=5x+3/4 ?

The solutions to the equation 3x^2+9/2 x-4=5x+3/4 are x=(1+sqrt(229))/(12),x=(1-sqrt(229))/(12)

Find the zeros of 3x^2+9/2 x-4=5x+3/4

The zeros of 3x^2+9/2 x-4=5x+3/4 are x=(1+sqrt(229))/(12),x=(1-sqrt(229))/(12)

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3 x^{2} + \frac{9}{2} x - 4 = 5 x + \frac{3}{4}

x = \frac{1 + 229}{12}, x = \frac{1 - 229}{12}

Decimal Notation

x = 1.34439 \dots, x = - 1.17772 \dots

Solution steps

3 x^{2} + \frac{9}{2} x - 4 = 5 x + \frac{3}{4}

Find Least Common Multiplier of

2, 4 : 4

Multiply by LCM=

4

3 x^{2} \cdot 4 + \frac{9}{2} x \cdot 4 - 4 \cdot 4 = 5 x \cdot 4 + \frac{3}{4} \cdot 4

Simplify

12 x^{2} + 18 x - 16 = 20 x + 3

Move

3

to the left side

12 x^{2} + 18 x - 19 = 20 x

Move

20 x

to the left side

12 x^{2} - 2 x - 19 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 12 ( - 19 )}{2 \cdot 12}

(- 2)^{2} - 4 \cdot 12 (- 19) = 2229

x_{1, 2} = \frac{- ( - 2 ) \pm 2 229}{2 \cdot 12}

Separate the solutions

x_{1} = \frac{- ( - 2 ) + 2 229}{2 \cdot 12}, x_{2} = \frac{- ( - 2 ) - 2 229}{2 \cdot 12}

x = \frac{- ( - 2 ) + 2 229}{2 \cdot 12} : \frac{1 + 229}{12}

x = \frac{- ( - 2 ) - 2 229}{2 \cdot 12} : \frac{1 - 229}{12}

The solutions to the quadratic equation are:

x = \frac{1 + 229}{12}, x = \frac{1 - 229}{12}

y, x + x y + 2 y^{2} = 6

2 x^{2} - 20 x + 58

3 x^{2} - 6 x + 9

(5 x)^{2} = 6

- 12 x^{2} - 6 x - 8 = 0

What are the solutions to the equation 3x^2+9/2 x-4=5x+3/4 ?
The solutions to the equation 3x^2+9/2 x-4=5x+3/4 are x=(1+sqrt(229))/(12),x=(1-sqrt(229))/(12)
Find the zeros of 3x^2+9/2 x-4=5x+3/4
The zeros of 3x^2+9/2 x-4=5x+3/4 are x=(1+sqrt(229))/(12),x=(1-sqrt(229))/(12)