What are the solutions to the equation 6x^2+48x+207=15 ?

The solutions to the equation 6x^2+48x+207=15 are x=-4+4i,x=-4-4i

Find the zeros of 6x^2+48x+207=15

The zeros of 6x^2+48x+207=15 are x=-4+4i,x=-4-4i

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6 x^{2} + 48 x + 207 = 15

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

Solution steps

6 x^{2} + 48 x + 207 = 15

Move

15

to the left side

6 x^{2} + 48 x + 192 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 48 \pm 4 8 ^{2} - 4 \cdot 6 \cdot 192}{2 \cdot 6}

Simplify

4 8^{2} - 4 \cdot 6 \cdot 192 : 48 i

x_{1, 2} = \frac{- 48 \pm 48 i}{2 \cdot 6}

Separate the solutions

x_{1} = \frac{- 48 + 48 i}{2 \cdot 6}, x_{2} = \frac{- 48 - 48 i}{2 \cdot 6}

x = \frac{- 48 + 48 i}{2 \cdot 6} : - 4 + 4 i

x = \frac{- 48 - 48 i}{2 \cdot 6} : - 4 - 4 i

The solutions to the quadratic equation are:

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

- 9 x^{2} + 81 = 0

2 r^{2} - 8 r = - 64

x^{2} + 5 x + 4

x^{2} + 5 x + 7

x^{2} = - 40

What are the solutions to the equation 6x^2+48x+207=15 ?
The solutions to the equation 6x^2+48x+207=15 are x=-4+4i,x=-4-4i
Find the zeros of 6x^2+48x+207=15
The zeros of 6x^2+48x+207=15 are x=-4+4i,x=-4-4i