What are the solutions to the equation 4x^2+14x+21=0 ?

The solutions to the equation 4x^2+14x+21=0 are x=-7/4+i(sqrt(35))/4 ,x=-7/4-i(sqrt(35))/4

Find the zeros of 4x^2+14x+21=0

The zeros of 4x^2+14x+21=0 are x=-7/4+i(sqrt(35))/4 ,x=-7/4-i(sqrt(35))/4

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4 x^{2} + 14 x + 21 = 0

x = - \frac{7}{4} + i \frac{35}{4}, x = - \frac{7}{4} - i \frac{35}{4}

Solution steps

4 x^{2} + 14 x + 21 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 14 \pm 1 4 ^{2} - 4 \cdot 4 \cdot 21}{2 \cdot 4}

Simplify

1 4^{2} - 4 \cdot 4 \cdot 21 : 235 i

x_{1, 2} = \frac{- 14 \pm 2 35 i}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- 14 + 2 35 i}{2 \cdot 4}, x_{2} = \frac{- 14 - 2 35 i}{2 \cdot 4}

x = \frac{- 14 + 2 35 i}{2 \cdot 4} : - \frac{7}{4} + i \frac{35}{4}

x = \frac{- 14 - 2 35 i}{2 \cdot 4} : - \frac{7}{4} - i \frac{35}{4}

The solutions to the quadratic equation are:

x = - \frac{7}{4} + i \frac{35}{4}, x = - \frac{7}{4} - i \frac{35}{4}

(4 x^{5} - 2 x^{3} + 2 x + 3) (4 x^{5} - 2 x^{3} + 2 x + 3)

j^{2} k^{2} + 3 j k^{2}

2 x + 7 - 2 (x - 1) = 3 (x + 3)

\frac{4 ^{2}}{4 ^{- 1}}

\frac{18 x ( x ^{3} - 2 )}{( x ^{3} + 1 ) ^{3}} \leq 0

What are the solutions to the equation 4x^2+14x+21=0 ?
The solutions to the equation 4x^2+14x+21=0 are x=-7/4+i(sqrt(35))/4 ,x=-7/4-i(sqrt(35))/4
Find the zeros of 4x^2+14x+21=0
The zeros of 4x^2+14x+21=0 are x=-7/4+i(sqrt(35))/4 ,x=-7/4-i(sqrt(35))/4