What are the solutions to the equation 3x^2+7=6x ?

The solutions to the equation 3x^2+7=6x are x=1+i(2sqrt(3))/3 ,x=1-i(2sqrt(3))/3

Find the zeros of 3x^2+7=6x

The zeros of 3x^2+7=6x are x=1+i(2sqrt(3))/3 ,x=1-i(2sqrt(3))/3

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3 x^{2} + 7 = 6 x

x = 1 + i \frac{2 3}{3}, x = 1 - i \frac{2 3}{3}

Solution steps

3 x^{2} + 7 = 6 x

Move

6 x

to the left side

3 x^{2} + 7 - 6 x = 0

Write in the standard form

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

3 x^{2} - 6 x + 7 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 6 ) \pm ( - 6 ) ^{2} - 4 \cdot 3 \cdot 7}{2 \cdot 3}

Simplify

(- 6)^{2} - 4 \cdot 3 \cdot 7 : 43 i

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

Separate the solutions

x_{1} = \frac{- ( - 6 ) + 4 3 i}{2 \cdot 3}, x_{2} = \frac{- ( - 6 ) - 4 3 i}{2 \cdot 3}

x = \frac{- ( - 6 ) + 4 3 i}{2 \cdot 3} : 1 + i \frac{2 3}{3}

x = \frac{- ( - 6 ) - 4 3 i}{2 \cdot 3} : 1 - i \frac{2 3}{3}

The solutions to the quadratic equation are:

x = 1 + i \frac{2 3}{3}, x = 1 - i \frac{2 3}{3}

\frac{2}{3} + \frac{1}{5} x - \frac{1}{4} = \frac{4}{5} x

- 3 x + 2 (x + 2) = 4 (- x - 3) - 1

y^{\frac{5}{2}}

p - 6 \leq - 3

x^{2} - 3 = 38

What are the solutions to the equation 3x^2+7=6x ?
The solutions to the equation 3x^2+7=6x are x=1+i(2sqrt(3))/3 ,x=1-i(2sqrt(3))/3
Find the zeros of 3x^2+7=6x
The zeros of 3x^2+7=6x are x=1+i(2sqrt(3))/3 ,x=1-i(2sqrt(3))/3