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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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 ( - 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}

4 x^{2} + 14 x - 8 = 0

- 3 x^{2} + 17 x = 10

3 x^{2} - 5 = 103

2 k^{2} = 22 - 4 k

(m + 3) x^{2} + 2 m = (2 m - 5) x + 10

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