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

The solutions to the equation 2x^2+9x+24=0 are x=-9/4+i(sqrt(111))/4 ,x=-9/4-i(sqrt(111))/4

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

The zeros of 2x^2+9x+24=0 are x=-9/4+i(sqrt(111))/4 ,x=-9/4-i(sqrt(111))/4

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

x = - \frac{9}{4} + i \frac{111}{4}, x = - \frac{9}{4} - i \frac{111}{4}

Solution steps

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

Solve with the quadratic formula

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

Simplify

9^{2} - 4 \cdot 2 \cdot 24 : 111 i

x_{1, 2} = \frac{- 9 \pm 111 i}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 9 + 111 i}{2 \cdot 2}, x_{2} = \frac{- 9 - 111 i}{2 \cdot 2}

x = \frac{- 9 + 111 i}{2 \cdot 2} : - \frac{9}{4} + i \frac{111}{4}

x = \frac{- 9 - 111 i}{2 \cdot 2} : - \frac{9}{4} - i \frac{111}{4}

The solutions to the quadratic equation are:

x = - \frac{9}{4} + i \frac{111}{4}, x = - \frac{9}{4} - i \frac{111}{4}

\frac{( x ^{2} y ) ^{- \frac{1}{3}} ( x y ) ^{\frac{1}{4}}}{( x y ^{- 2} ) ^{\frac{1}{12}}}

1 \cdot \frac{1}{197.927} \cdot \frac{2}{4} \cdot 159.69

2 x^{3} + 18 x^{2} + 66 x + 50

4 x^{3} - 12 y^{4}

x^{2} - 8 x + 12 \leq 0

What are the solutions to the equation 2x^2+9x+24=0 ?
The solutions to the equation 2x^2+9x+24=0 are x=-9/4+i(sqrt(111))/4 ,x=-9/4-i(sqrt(111))/4
Find the zeros of 2x^2+9x+24=0
The zeros of 2x^2+9x+24=0 are x=-9/4+i(sqrt(111))/4 ,x=-9/4-i(sqrt(111))/4