What are the solutions to the equation 9x^2+24x+19=0 ?

The solutions to the equation 9x^2+24x+19=0 are x=-4/3+i(sqrt(3))/3 ,x=-4/3-i(sqrt(3))/3

Find the zeros of 9x^2+24x+19=0

The zeros of 9x^2+24x+19=0 are x=-4/3+i(sqrt(3))/3 ,x=-4/3-i(sqrt(3))/3

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

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

Solution steps

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

Solve with the quadratic formula

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

Simplify

2 4^{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}

4 x^{2} + 10 x - 3 = 0

t, d = 4.9 t^{2}

u^{2} - 6 u + 9 = 0

x (x - 5) = - 6

x^{2} + 17 x + 3 = 0

What are the solutions to the equation 9x^2+24x+19=0 ?
The solutions to the equation 9x^2+24x+19=0 are x=-4/3+i(sqrt(3))/3 ,x=-4/3-i(sqrt(3))/3
Find the zeros of 9x^2+24x+19=0
The zeros of 9x^2+24x+19=0 are x=-4/3+i(sqrt(3))/3 ,x=-4/3-i(sqrt(3))/3