What are the solutions to the equation 3x^2+0.6x=16.8 ?

The solutions to the equation 3x^2+0.6x=16.8 are x=(-1+sqrt(561))/(10),x=-(1+sqrt(561))/(10)

Find the zeros of 3x^2+0.6x=16.8

The zeros of 3x^2+0.6x=16.8 are x=(-1+sqrt(561))/(10),x=-(1+sqrt(561))/(10)

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3 x^{2} + 0.6 x = 16.8

x = \frac{- 1 + 561}{10}, x = - \frac{1 + 561}{10}

Decimal Notation

x = 2.26854 \dots, x = - 2.46854 \dots

Solution steps

3 x^{2} + 0.6 x = 16.8

Multiply both sides by

10

30 x^{2} + 6 x = 168

Move

168

to the left side

30 x^{2} + 6 x - 168 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 6 \pm 6 ^{2} - 4 \cdot 30 ( - 168 )}{2 \cdot 30}

6^{2} - 4 \cdot 30 (- 168) = 6561

x_{1, 2} = \frac{- 6 \pm 6 561}{2 \cdot 30}

Separate the solutions

x_{1} = \frac{- 6 + 6 561}{2 \cdot 30}, x_{2} = \frac{- 6 - 6 561}{2 \cdot 30}

x = \frac{- 6 + 6 561}{2 \cdot 30} : \frac{- 1 + 561}{10}

x = \frac{- 6 - 6 561}{2 \cdot 30} : - \frac{1 + 561}{10}

The solutions to the quadratic equation are:

x = \frac{- 1 + 561}{10}, x = - \frac{1 + 561}{10}

3 x^{2} - x = - 4 x + 60

- \frac{1}{4} + \frac{3}{4}

v^{2} + 10 v + 25 = 0

- 16 t^{2} + 36 t - 20

216 y^{2}

What are the solutions to the equation 3x^2+0.6x=16.8 ?
The solutions to the equation 3x^2+0.6x=16.8 are x=(-1+sqrt(561))/(10),x=-(1+sqrt(561))/(10)
Find the zeros of 3x^2+0.6x=16.8
The zeros of 3x^2+0.6x=16.8 are x=(-1+sqrt(561))/(10),x=-(1+sqrt(561))/(10)