What are the solutions to the equation 3x^2+24x+48=-108 ?

The solutions to the equation 3x^2+24x+48=-108 are x=-4+6i,x=-4-6i

Find the zeros of 3x^2+24x+48=-108

The zeros of 3x^2+24x+48=-108 are x=-4+6i,x=-4-6i

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3 x^{2} + 24 x + 48 = - 108

x = - 4 + 6 i, x = - 4 - 6 i

Solution steps

3 x^{2} + 24 x + 48 = - 108

Move

108

to the left side

3 x^{2} + 24 x + 156 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 24 \pm 2 4 ^{2} - 4 \cdot 3 \cdot 156}{2 \cdot 3}

Simplify

2 4^{2} - 4 \cdot 3 \cdot 156 : 36 i

x_{1, 2} = \frac{- 24 \pm 36 i}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- 24 + 36 i}{2 \cdot 3}, x_{2} = \frac{- 24 - 36 i}{2 \cdot 3}

x = \frac{- 24 + 36 i}{2 \cdot 3} : - 4 + 6 i

x = \frac{- 24 - 36 i}{2 \cdot 3} : - 4 - 6 i

The solutions to the quadratic equation are:

x = - 4 + 6 i, x = - 4 - 6 i

2 x^{2} + 3 x - 20 = 0

x^{2} + \frac{3}{2} x - 3 = 0

2 x^{2} - 9 x + 21 = 0

y, y^{2} - 3 x y + 5 x^{2} y = 13 + x y^{2}

x^{2} + 10 x + 21 = 0

What are the solutions to the equation 3x^2+24x+48=-108 ?
The solutions to the equation 3x^2+24x+48=-108 are x=-4+6i,x=-4-6i
Find the zeros of 3x^2+24x+48=-108
The zeros of 3x^2+24x+48=-108 are x=-4+6i,x=-4-6i