What are the solutions to the equation 9x^2+19=24x ?

The solutions to the equation 9x^2+19=24x are x= 4/3+i(sqrt(3))/3 ,x= 4/3-i(sqrt(3))/3

Find the zeros of 9x^2+19=24x

The zeros of 9x^2+19=24x are x= 4/3+i(sqrt(3))/3 ,x= 4/3-i(sqrt(3))/3

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9 x^{2} + 19 = 24 x

x = \frac{4}{3} + i \frac{3}{3}, x = \frac{4}{3} - i \frac{3}{3}

Solution steps

9 x^{2} + 19 = 24 x

Move

24 x

to the left side

9 x^{2} + 19 - 24 x = 0

Write in the standard form

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

9 x^{2} - 24 x + 19 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 24 ) \pm ( - 24 ) ^{2} - 4 \cdot 9 \cdot 19}{2 \cdot 9}

Simplify

(- 24)^{2} - 4 \cdot 9 \cdot 19 : 63 i

x_{1, 2} = \frac{- ( - 24 ) \pm 6 3 i}{2 \cdot 9}

Separate the solutions

x_{1} = \frac{- ( - 24 ) + 6 3 i}{2 \cdot 9}, x_{2} = \frac{- ( - 24 ) - 6 3 i}{2 \cdot 9}

x = \frac{- ( - 24 ) + 6 3 i}{2 \cdot 9} : \frac{4}{3} + i \frac{3}{3}

x = \frac{- ( - 24 ) - 6 3 i}{2 \cdot 9} : \frac{4}{3} - i \frac{3}{3}

The solutions to the quadratic equation are:

x = \frac{4}{3} + i \frac{3}{3}, x = \frac{4}{3} - i \frac{3}{3}

- 3 x^{2} + 5 x - 11 = 0

n^{2} + 11 n + 3 = - 4 n^{2}

20 - x^{2} = 10

x^{2} - 7 x + \frac{1}{4} = 0

x^{2} + 25 = - 24

What are the solutions to the equation 9x^2+19=24x ?
The solutions to the equation 9x^2+19=24x are x= 4/3+i(sqrt(3))/3 ,x= 4/3-i(sqrt(3))/3
Find the zeros of 9x^2+19=24x
The zeros of 9x^2+19=24x are x= 4/3+i(sqrt(3))/3 ,x= 4/3-i(sqrt(3))/3