What are the solutions to the equation 1+(5x)/6 =(2x^2)/3 ?

The solutions to the equation 1+(5x)/6 =(2x^2)/3 are x=2,x=-3/4

Find the zeros of 1+(5x)/6 =(2x^2)/3

The zeros of 1+(5x)/6 =(2x^2)/3 are x=2,x=-3/4

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1 + \frac{5 x}{6} = \frac{2 x ^{2}}{3}

x = 2, x = - \frac{3}{4}

Decimal Notation

x = 2, x = - 0.75

Solution steps

1 + \frac{5 x}{6} = \frac{2 x ^{2}}{3}

Find Least Common Multiplier of

6, 3 : 6

Multiply by LCM=

6

1 \cdot 6 + \frac{5 x}{6} \cdot 6 = \frac{2 x ^{2}}{3} \cdot 6

Simplify

6 + 5 x = 4 x^{2}

Switch sides

4 x^{2} = 6 + 5 x

Move

5 x

to the left side

4 x^{2} - 5 x = 6

Move

6

to the left side

4 x^{2} - 5 x - 6 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 4 ( - 6 )}{2 \cdot 4}

(- 5)^{2} - 4 \cdot 4 (- 6) = 11

x_{1, 2} = \frac{- ( - 5 ) \pm 11}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- ( - 5 ) + 11}{2 \cdot 4}, x_{2} = \frac{- ( - 5 ) - 11}{2 \cdot 4}

x = \frac{- ( - 5 ) + 11}{2 \cdot 4} : 2

x = \frac{- ( - 5 ) - 11}{2 \cdot 4} : - \frac{3}{4}

The solutions to the quadratic equation are:

x = 2, x = - \frac{3}{4}

8 x^{2} - 4 x - 1 = 0

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

x^{2} + 15 x + 30 = 3 x - 23 x - 2

4 m^{2} + 5 m = 7

(x^{2} + 3 x - 2.5) = 0

What are the solutions to the equation 1+(5x)/6 =(2x^2)/3 ?
The solutions to the equation 1+(5x)/6 =(2x^2)/3 are x=2,x=-3/4
Find the zeros of 1+(5x)/6 =(2x^2)/3
The zeros of 1+(5x)/6 =(2x^2)/3 are x=2,x=-3/4