What are the solutions to the equation 0=x^2+3x+4 ?

The solutions to the equation 0=x^2+3x+4 are x=-3/2+i(sqrt(7))/2 ,x=-3/2-i(sqrt(7))/2

Find the zeros of 0=x^2+3x+4

The zeros of 0=x^2+3x+4 are x=-3/2+i(sqrt(7))/2 ,x=-3/2-i(sqrt(7))/2

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

x = - \frac{3}{2} + i \frac{7}{2}, x = - \frac{3}{2} - i \frac{7}{2}

Solution steps

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

Switch sides

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

Solve with the quadratic formula

x_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 \cdot 1 \cdot 4}{2 \cdot 1}

Simplify

3^{2} - 4 \cdot 1 \cdot 4 : 7 i

x_{1, 2} = \frac{- 3 \pm 7 i}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 3 + 7 i}{2 \cdot 1}, x_{2} = \frac{- 3 - 7 i}{2 \cdot 1}

x = \frac{- 3 + 7 i}{2 \cdot 1} : - \frac{3}{2} + i \frac{7}{2}

x = \frac{- 3 - 7 i}{2 \cdot 1} : - \frac{3}{2} - i \frac{7}{2}

The solutions to the quadratic equation are:

x = - \frac{3}{2} + i \frac{7}{2}, x = - \frac{3}{2} - i \frac{7}{2}

y^{2} - 14 y + 40 = 0

(1 - x)^{2} = \frac{9}{25}

x^{2} + 7 x = 0

(z - 4)^{2} - 12 (z - 4) + 27 = 0

x^{2} - \frac{1}{3} = \frac{1}{6} x^{2}

What are the solutions to the equation 0=x^2+3x+4 ?
The solutions to the equation 0=x^2+3x+4 are x=-3/2+i(sqrt(7))/2 ,x=-3/2-i(sqrt(7))/2
Find the zeros of 0=x^2+3x+4
The zeros of 0=x^2+3x+4 are x=-3/2+i(sqrt(7))/2 ,x=-3/2-i(sqrt(7))/2