What are the solutions to the equation 2x^2+20x+57=4 ?

The solutions to the equation 2x^2+20x+57=4 are x=-5+isqrt(3/2),x=-5-isqrt(3/2)

Find the zeros of 2x^2+20x+57=4

The zeros of 2x^2+20x+57=4 are x=-5+isqrt(3/2),x=-5-isqrt(3/2)

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2 x^{2} + 20 x + 57 = 4

x = - 5 + i \frac{3}{2}, x = - 5 - i \frac{3}{2}

Solution steps

2 x^{2} + 20 x + 57 = 4

Move

4

to the left side

2 x^{2} + 20 x + 53 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 20 \pm 2 0 ^{2} - 4 \cdot 2 \cdot 53}{2 \cdot 2}

Simplify

2 0^{2} - 4 \cdot 2 \cdot 53 : 26 i

x_{1, 2} = \frac{- 20 \pm 2 6 i}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 20 + 2 6 i}{2 \cdot 2}, x_{2} = \frac{- 20 - 2 6 i}{2 \cdot 2}

x = \frac{- 20 + 2 6 i}{2 \cdot 2} : - 5 + i \frac{3}{2}

x = \frac{- 20 - 2 6 i}{2 \cdot 2} : - 5 - i \frac{3}{2}

The solutions to the quadratic equation are:

x = - 5 + i \frac{3}{2}, x = - 5 - i \frac{3}{2}

5 x - 4 = x + 3

3 n^{2} + n - 80 = 0

5 (x + 1) = 4 x + 7

\frac{4}{15} - \frac{4}{5} + \frac{1}{2}

3 x^{2} - 23 x y + 40 y^{2}

What are the solutions to the equation 2x^2+20x+57=4 ?
The solutions to the equation 2x^2+20x+57=4 are x=-5+isqrt(3/2),x=-5-isqrt(3/2)
Find the zeros of 2x^2+20x+57=4
The zeros of 2x^2+20x+57=4 are x=-5+isqrt(3/2),x=-5-isqrt(3/2)