What are the solutions to the equation x^2+32x+116=12x ?

The solutions to the equation x^2+32x+116=12x are x=-10+4i,x=-10-4i

Find the zeros of x^2+32x+116=12x

The zeros of x^2+32x+116=12x are x=-10+4i,x=-10-4i

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x^{2} + 32 x + 116 = 12 x

x = - 10 + 4 i, x = - 10 - 4 i

Solution steps

x^{2} + 32 x + 116 = 12 x

Move

12 x

to the left side

x^{2} + 20 x + 116 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 20 \pm 2 0 ^{2} - 4 \cdot 1 \cdot 116}{2 \cdot 1}

Simplify

2 0^{2} - 4 \cdot 1 \cdot 116 : 8 i

x_{1, 2} = \frac{- 20 \pm 8 i}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 20 + 8 i}{2 \cdot 1}, x_{2} = \frac{- 20 - 8 i}{2 \cdot 1}

x = \frac{- 20 + 8 i}{2 \cdot 1} : - 10 + 4 i

x = \frac{- 20 - 8 i}{2 \cdot 1} : - 10 - 4 i

The solutions to the quadratic equation are:

x = - 10 + 4 i, x = - 10 - 4 i

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

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

y, - 16 y^{2} + 9 z^{2} = 0

125 x^{2} + 50 x - 5 = 0

t, x = t^{2} + t

What are the solutions to the equation x^2+32x+116=12x ?
The solutions to the equation x^2+32x+116=12x are x=-10+4i,x=-10-4i
Find the zeros of x^2+32x+116=12x
The zeros of x^2+32x+116=12x are x=-10+4i,x=-10-4i