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

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

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

The zeros of 2x^2+bx=3 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

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

Solution steps

2 x^{2} + b x = 3

Move

3

to the left side

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

Solve with the quadratic formula

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

Simplify

b^{2} - 4 \cdot 2 (- 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}

- 0.5358 x^{2} - 7.2142 x + 184.57 = 0

3 x^{2} - 8 x = 0

x = - 2 x^{2} + 8 x - \frac{7}{2}

25 x^{2} - 1 = 0

x^{2} - 16 x + 64 = - 8

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