What are the solutions to the equation 4x^2+32x=-68 ?

The solutions to the equation 4x^2+32x=-68 are x=-4+i,x=-4-i

Find the zeros of 4x^2+32x=-68

The zeros of 4x^2+32x=-68 are x=-4+i,x=-4-i

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4 x^{2} + 32 x = - 68

x = - 4 + i, x = - 4 - i

Solution steps

4 x^{2} + 32 x = - 68

Move

68

to the left side

4 x^{2} + 32 x + 68 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 32 \pm 3 2 ^{2} - 4 \cdot 4 \cdot 68}{2 \cdot 4}

Simplify

3 2^{2} - 4 \cdot 4 \cdot 68 : 8 i

x_{1, 2} = \frac{- 32 \pm 8 i}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- 32 + 8 i}{2 \cdot 4}, x_{2} = \frac{- 32 - 8 i}{2 \cdot 4}

x = \frac{- 32 + 8 i}{2 \cdot 4} : - 4 + i

x = \frac{- 32 - 8 i}{2 \cdot 4} : - 4 - i

The solutions to the quadratic equation are:

x = - 4 + i, x = - 4 - i

y, x^{2} - x y^{2} = 2

4 x^{2} + 21 = 4 x

x^{2} + 5 x + 4

x^{2} - 49

x^{2} + 2 x + 1 = 0

What are the solutions to the equation 4x^2+32x=-68 ?
The solutions to the equation 4x^2+32x=-68 are x=-4+i,x=-4-i
Find the zeros of 4x^2+32x=-68
The zeros of 4x^2+32x=-68 are x=-4+i,x=-4-i