What are the solutions to the equation 3x^2+6=19x ?

The solutions to the equation 3x^2+6=19x are x=6,x= 1/3

Find the zeros of 3x^2+6=19x

The zeros of 3x^2+6=19x are x=6,x= 1/3

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3 x^{2} + 6 = 19 x

x = 6, x = \frac{1}{3}

Decimal Notation

x = 6, x = 0.33333 \dots

Solution steps

3 x^{2} + 6 = 19 x

Move

19 x

to the left side

3 x^{2} + 6 - 19 x = 0

Write in the standard form

a x^{2} + b x + c = 0

3 x^{2} - 19 x + 6 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 19 ) \pm ( - 19 ) ^{2} - 4 \cdot 3 \cdot 6}{2 \cdot 3}

(- 19)^{2} - 4 \cdot 3 \cdot 6 = 17

x_{1, 2} = \frac{- ( - 19 ) \pm 17}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- ( - 19 ) + 17}{2 \cdot 3}, x_{2} = \frac{- ( - 19 ) - 17}{2 \cdot 3}

x = \frac{- ( - 19 ) + 17}{2 \cdot 3} : 6

x = \frac{- ( - 19 ) - 17}{2 \cdot 3} : \frac{1}{3}

The solutions to the quadratic equation are:

x = 6, x = \frac{1}{3}

- 4 \cdot \frac{1}{4}

\frac{2 x ^{2} - 11 x + 9}{9 x - 9}

- 5 \leq x

- \frac{1}{2} \cdot 5

\frac{5}{( 2 ( \frac{1}{3} ) + 1 ) ^{2}}

What are the solutions to the equation 3x^2+6=19x ?
The solutions to the equation 3x^2+6=19x are x=6,x= 1/3
Find the zeros of 3x^2+6=19x
The zeros of 3x^2+6=19x are x=6,x= 1/3