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

The solutions to the equation 2x^2+bx+2=0 are x=(-b+sqrt(b^2-16))/4 ,x=(-b-sqrt(b^2-16))/4

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

The zeros of 2x^2+bx+2=0 are x=(-b+sqrt(b^2-16))/4 ,x=(-b-sqrt(b^2-16))/4

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

x = \frac{- b + b ^{2} - 16}{4}, x = \frac{- b - b ^{2} - 16}{4}

Solution steps

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

Solve with the quadratic formula

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

Simplify

b^{2} - 4 \cdot 2 \cdot 2 : b^{2} - 16

x_{1, 2} = \frac{- b \pm b ^{2} - 16}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- b + b ^{2} - 16}{2 \cdot 2}, x_{2} = \frac{- b - b ^{2} - 16}{2 \cdot 2}

x = \frac{- b + b ^{2} - 16}{2 \cdot 2} : \frac{- b + b ^{2} - 16}{4}

x = \frac{- b - b ^{2} - 16}{2 \cdot 2} : \frac{- b - b ^{2} - 16}{4}

The solutions to the quadratic equation are:

x = \frac{- b + b ^{2} - 16}{4}, x = \frac{- b - b ^{2} - 16}{4}

2 y^{2} - 8 = 6 - 2 y^{2}

x, \frac{x ^{2}}{18} = R

x, x^{2} y = 9

9 x^{2} - 30 x + 23 = 0

m^{2} + 12 = 48

What are the solutions to the equation 2x^2+bx+2=0 ?
The solutions to the equation 2x^2+bx+2=0 are x=(-b+sqrt(b^2-16))/4 ,x=(-b-sqrt(b^2-16))/4
Find the zeros of 2x^2+bx+2=0
The zeros of 2x^2+bx+2=0 are x=(-b+sqrt(b^2-16))/4 ,x=(-b-sqrt(b^2-16))/4