What are the solutions to the equation 2x^2+3x+c=0 ?

The solutions to the equation 2x^2+3x+c=0 are x=(-3+sqrt(9-8c))/4 ,x=(-3-sqrt(9-8c))/4

Find the zeros of 2x^2+3x+c=0

The zeros of 2x^2+3x+c=0 are x=(-3+sqrt(9-8c))/4 ,x=(-3-sqrt(9-8c))/4

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2 x^{2} + 3 x + c = 0

x = \frac{- 3 + 9 - 8 c}{4}, x = \frac{- 3 - 9 - 8 c}{4}

Solution steps

2 x^{2} + 3 x + c = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 \cdot 2 c}{2 \cdot 2}

Simplify

3^{2} - 4 \cdot 2 c : 9 - 8 c

x_{1, 2} = \frac{- 3 \pm 9 - 8 c}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 3 + 9 - 8 c}{2 \cdot 2}, x_{2} = \frac{- 3 - 9 - 8 c}{2 \cdot 2}

x = \frac{- 3 + 9 - 8 c}{2 \cdot 2} : \frac{- 3 + 9 - 8 c}{4}

x = \frac{- 3 - 9 - 8 c}{2 \cdot 2} : \frac{- 3 - 9 - 8 c}{4}

The solutions to the quadratic equation are:

x = \frac{- 3 + 9 - 8 c}{4}, x = \frac{- 3 - 9 - 8 c}{4}

a^{2} + 5 a + 4 = 0

x, x^{2} + x - p (p + 1) = 0

(7) \cdot 2 \cdot x^{2} + 7 x + 52 = 0

x, z = x y + f (x^{2} + y^{2})

h, r^{2} = h^{2} + \frac{p ^{2}}{q ^{2}}

What are the solutions to the equation 2x^2+3x+c=0 ?
The solutions to the equation 2x^2+3x+c=0 are x=(-3+sqrt(9-8c))/4 ,x=(-3-sqrt(9-8c))/4
Find the zeros of 2x^2+3x+c=0
The zeros of 2x^2+3x+c=0 are x=(-3+sqrt(9-8c))/4 ,x=(-3-sqrt(9-8c))/4