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

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

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

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

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x^{2} + 3 x = - 4

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

Solution steps

x^{2} + 3 x = - 4

Move

4

to the left side

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}

2 x^{2} - 250 x + 5000 = 0

62.5 = 10 (b)^{2}

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

b, A = \frac{1}{2} h (b^{1} + b^{2})

2 x^{2} - 8 x - 8 = 0

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