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

The solutions to the equation 2x^2+bx+3=0 are x=(-b+sqrt(b^2-24))/4 ,x=(-b-sqrt(b^2-24))/4

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

The zeros of 2x^2+bx+3=0 are x=(-b+sqrt(b^2-24))/4 ,x=(-b-sqrt(b^2-24))/4

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

x = \frac{- b + b ^{2} - 24}{4}, x = \frac{- b - b ^{2} - 24}{4}

Solution steps

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

Solve with the quadratic formula

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

Simplify

b^{2} - 4 \cdot 2 \cdot 3 : b^{2} - 24

x_{1, 2} = \frac{- b \pm b ^{2} - 24}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- b + b ^{2} - 24}{2 \cdot 2}, x_{2} = \frac{- b - b ^{2} - 24}{2 \cdot 2}

x = \frac{- b + b ^{2} - 24}{2 \cdot 2} : \frac{- b + b ^{2} - 24}{4}

x = \frac{- b - b ^{2} - 24}{2 \cdot 2} : \frac{- b - b ^{2} - 24}{4}

The solutions to the quadratic equation are:

x = \frac{- b + b ^{2} - 24}{4}, x = \frac{- b - b ^{2} - 24}{4}

x^{2} + 5.71 \cdot 1 0^{- 12} x - 5.71 \cdot 1 0^{- 10} = 0

a, x^{2} + y^{2} - x y = a^{2}

y^{2} - 2 y + 1 = y + 1

D, (D^{2} - \frac{3}{x} D + \frac{3}{x ^{2}}) y = 2 x - 1

x^{2} + 96 = - 19 x

What are the solutions to the equation 2x^2+bx+3=0 ?
The solutions to the equation 2x^2+bx+3=0 are x=(-b+sqrt(b^2-24))/4 ,x=(-b-sqrt(b^2-24))/4
Find the zeros of 2x^2+bx+3=0
The zeros of 2x^2+bx+3=0 are x=(-b+sqrt(b^2-24))/4 ,x=(-b-sqrt(b^2-24))/4