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

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

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

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

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

x = \frac{- b + b ^{2} - 8 c}{4}, x = \frac{- b - b ^{2} - 8 c}{4}

Solution steps

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

Solve with the quadratic formula

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

Simplify

b^{2} - 4 \cdot 2 c : b^{2} - 8 c

x_{1, 2} = \frac{- b \pm b ^{2} - 8 c}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- b + b ^{2} - 8 c}{2 \cdot 2}, x_{2} = \frac{- b - b ^{2} - 8 c}{2 \cdot 2}

x = \frac{- b + b ^{2} - 8 c}{2 \cdot 2} : \frac{- b + b ^{2} - 8 c}{4}

x = \frac{- b - b ^{2} - 8 c}{2 \cdot 2} : \frac{- b - b ^{2} - 8 c}{4}

The solutions to the quadratic equation are:

x = \frac{- b + b ^{2} - 8 c}{4}, x = \frac{- b - b ^{2} - 8 c}{4}

x^{2} - 2 x - 4 = 0

6 (3 x - 1) (2 x - 1) - 12 x^{2} = 7 x (2 x - 1)

- x^{2} + 3 x + 4 = x + 1

8 x^{2} - x + 1 = 0

x^{2} = 30 + x

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