What are the solutions to the equation 3x^2+39=18x ?

The solutions to the equation 3x^2+39=18x are x=3+2i,x=3-2i

Find the zeros of 3x^2+39=18x

The zeros of 3x^2+39=18x are x=3+2i,x=3-2i

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3 x^{2} + 39 = 18 x

x = 3 + 2 i, x = 3 - 2 i

Solution steps

3 x^{2} + 39 = 18 x

Move

18 x

to the left side

3 x^{2} + 39 - 18 x = 0

Write in the standard form

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

3 x^{2} - 18 x + 39 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 18 ) \pm ( - 18 ) ^{2} - 4 \cdot 3 \cdot 39}{2 \cdot 3}

Simplify

(- 18)^{2} - 4 \cdot 3 \cdot 39 : 12 i

x_{1, 2} = \frac{- ( - 18 ) \pm 12 i}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- ( - 18 ) + 12 i}{2 \cdot 3}, x_{2} = \frac{- ( - 18 ) - 12 i}{2 \cdot 3}

x = \frac{- ( - 18 ) + 12 i}{2 \cdot 3} : 3 + 2 i

x = \frac{- ( - 18 ) - 12 i}{2 \cdot 3} : 3 - 2 i

The solutions to the quadratic equation are:

x = 3 + 2 i, x = 3 - 2 i

2^{\frac{1}{3}} \cdot 4^{\frac{1}{3}}

12 = a^{2}

4 (2 x - 4) - 5 x + 4 = - 30

s^{2} - s - 12 = 0

0 \leq 4 - \frac{1}{3} x \leq 2

What are the solutions to the equation 3x^2+39=18x ?
The solutions to the equation 3x^2+39=18x are x=3+2i,x=3-2i
Find the zeros of 3x^2+39=18x
The zeros of 3x^2+39=18x are x=3+2i,x=3-2i