What are the solutions to the equation 8x^2+4x=-5 ?

The solutions to the equation 8x^2+4x=-5 are x=-1/4+i 3/4 ,x=-1/4-i 3/4

Find the zeros of 8x^2+4x=-5

The zeros of 8x^2+4x=-5 are x=-1/4+i 3/4 ,x=-1/4-i 3/4

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8 x^{2} + 4 x = - 5

x = - \frac{1}{4} + i \frac{3}{4}, x = - \frac{1}{4} - i \frac{3}{4}

Solution steps

8 x^{2} + 4 x = - 5

Move

5

to the left side

8 x^{2} + 4 x + 5 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 8 \cdot 5}{2 \cdot 8}

Simplify

4^{2} - 4 \cdot 8 \cdot 5 : 12 i

x_{1, 2} = \frac{- 4 \pm 12 i}{2 \cdot 8}

Separate the solutions

x_{1} = \frac{- 4 + 12 i}{2 \cdot 8}, x_{2} = \frac{- 4 - 12 i}{2 \cdot 8}

x = \frac{- 4 + 12 i}{2 \cdot 8} : - \frac{1}{4} + i \frac{3}{4}

x = \frac{- 4 - 12 i}{2 \cdot 8} : - \frac{1}{4} - i \frac{3}{4}

The solutions to the quadratic equation are:

x = - \frac{1}{4} + i \frac{3}{4}, x = - \frac{1}{4} - i \frac{3}{4}

(4 + k) x^{2} - 2 k x + 2 = 0

45 x^{2} - 36 = 0

x (3 x + 6) = 140

x, f (x + 1) = (x + 1)^{2} - 4

11 x^{2} - 12 x = 0

What are the solutions to the equation 8x^2+4x=-5 ?
The solutions to the equation 8x^2+4x=-5 are x=-1/4+i 3/4 ,x=-1/4-i 3/4
Find the zeros of 8x^2+4x=-5
The zeros of 8x^2+4x=-5 are x=-1/4+i 3/4 ,x=-1/4-i 3/4