What are the solutions to the equation 8x^2+36x+24=12x+5 ?

The solutions to the equation 8x^2+36x+24=12x+5 are x=-3/2+i(sqrt(2))/4 ,x=-3/2-i(sqrt(2))/4

Find the zeros of 8x^2+36x+24=12x+5

The zeros of 8x^2+36x+24=12x+5 are x=-3/2+i(sqrt(2))/4 ,x=-3/2-i(sqrt(2))/4

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8 x^{2} + 36 x + 24 = 12 x + 5

x = - \frac{3}{2} + i \frac{2}{4}, x = - \frac{3}{2} - i \frac{2}{4}

Solution steps

8 x^{2} + 36 x + 24 = 12 x + 5

Move

5

to the left side

8 x^{2} + 36 x + 19 = 12 x

Move

12 x

to the left side

8 x^{2} + 24 x + 19 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 24 \pm 2 4 ^{2} - 4 \cdot 8 \cdot 19}{2 \cdot 8}

Simplify

2 4^{2} - 4 \cdot 8 \cdot 19 : 42 i

x_{1, 2} = \frac{- 24 \pm 4 2 i}{2 \cdot 8}

Separate the solutions

x_{1} = \frac{- 24 + 4 2 i}{2 \cdot 8}, x_{2} = \frac{- 24 - 4 2 i}{2 \cdot 8}

x = \frac{- 24 + 4 2 i}{2 \cdot 8} : - \frac{3}{2} + i \frac{2}{4}

x = \frac{- 24 - 4 2 i}{2 \cdot 8} : - \frac{3}{2} - i \frac{2}{4}

The solutions to the quadratic equation are:

x = - \frac{3}{2} + i \frac{2}{4}, x = - \frac{3}{2} - i \frac{2}{4}

\frac{2}{3} \cdot \frac{15}{14}

\frac{1}{3} (12)

3 x - 3 = x + 4

x^{2} + 13 x y - 14 y^{2}

6 (2 x + 4) = 3 (4 x + 8)

What are the solutions to the equation 8x^2+36x+24=12x+5 ?
The solutions to the equation 8x^2+36x+24=12x+5 are x=-3/2+i(sqrt(2))/4 ,x=-3/2-i(sqrt(2))/4
Find the zeros of 8x^2+36x+24=12x+5
The zeros of 8x^2+36x+24=12x+5 are x=-3/2+i(sqrt(2))/4 ,x=-3/2-i(sqrt(2))/4