What are the solutions to the equation (6x^2+2x)/3 =20x ?

The solutions to the equation (6x^2+2x)/3 =20x are x= 29/3 ,x=0

Find the zeros of (6x^2+2x)/3 =20x

The zeros of (6x^2+2x)/3 =20x are x= 29/3 ,x=0

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\frac{6 x ^{2} + 2 x}{3} = 20 x

x = \frac{29}{3}, x = 0

Decimal Notation

x = 9.66666 \dots, x = 0

Solution steps

\frac{6 x ^{2} + 2 x}{3} = 20 x

Multiply both sides by

3

6 x^{2} + 2 x = 60 x

Move

60 x

to the left side

6 x^{2} - 58 x = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 58 ) \pm ( - 58 ) ^{2} - 4 \cdot 6 \cdot 0}{2 \cdot 6}

(- 58)^{2} - 4 \cdot 6 \cdot 0 = 58

x_{1, 2} = \frac{- ( - 58 ) \pm 58}{2 \cdot 6}

Separate the solutions

x_{1} = \frac{- ( - 58 ) + 58}{2 \cdot 6}, x_{2} = \frac{- ( - 58 ) - 58}{2 \cdot 6}

x = \frac{- ( - 58 ) + 58}{2 \cdot 6} : \frac{29}{3}

x = \frac{- ( - 58 ) - 58}{2 \cdot 6} : 0

The solutions to the quadratic equation are:

x = \frac{29}{3}, x = 0

x^{2} - 9 x = - 29

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

- 9 x^{2} + 153 x - 144 = 0

2 x^{2} + 4 x + 30 = 0

x^{2} + 2 x - 0.5 = 0

What are the solutions to the equation (6x^2+2x)/3 =20x ?
The solutions to the equation (6x^2+2x)/3 =20x are x= 29/3 ,x=0
Find the zeros of (6x^2+2x)/3 =20x
The zeros of (6x^2+2x)/3 =20x are x= 29/3 ,x=0