What are the solutions to the equation 2x^2+5x+4=0 ?

The solutions to the equation 2x^2+5x+4=0 are x=-5/4+i(sqrt(7))/4 ,x=-5/4-i(sqrt(7))/4

Find the zeros of 2x^2+5x+4=0

The zeros of 2x^2+5x+4=0 are x=-5/4+i(sqrt(7))/4 ,x=-5/4-i(sqrt(7))/4

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2 x^{2} + 5 x + 4 = 0

x = - \frac{5}{4} + i \frac{7}{4}, x = - \frac{5}{4} - i \frac{7}{4}

Solution steps

2 x^{2} + 5 x + 4 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 \cdot 2 \cdot 4}{2 \cdot 2}

Simplify

5^{2} - 4 \cdot 2 \cdot 4 : 7 i

x_{1, 2} = \frac{- 5 \pm 7 i}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 5 + 7 i}{2 \cdot 2}, x_{2} = \frac{- 5 - 7 i}{2 \cdot 2}

x = \frac{- 5 + 7 i}{2 \cdot 2} : - \frac{5}{4} + i \frac{7}{4}

x = \frac{- 5 - 7 i}{2 \cdot 2} : - \frac{5}{4} - i \frac{7}{4}

The solutions to the quadratic equation are:

x = - \frac{5}{4} + i \frac{7}{4}, x = - \frac{5}{4} - i \frac{7}{4}

- 9 g + 9 > - g - 7

2 t + (t - 4)^{2} = 5 t (t - 4) + 24

(\frac{3}{5})^{- 4}

\frac{3 ^{- 8}}{3 ^{- 4}}

6 k^{2} + 2 k + 9 = - 3

What are the solutions to the equation 2x^2+5x+4=0 ?
The solutions to the equation 2x^2+5x+4=0 are x=-5/4+i(sqrt(7))/4 ,x=-5/4-i(sqrt(7))/4
Find the zeros of 2x^2+5x+4=0
The zeros of 2x^2+5x+4=0 are x=-5/4+i(sqrt(7))/4 ,x=-5/4-i(sqrt(7))/4