What are the solutions to the equation 4x^2+12x-15=0 ?

The solutions to the equation 4x^2+12x-15=0 are x=(-3+2sqrt(6))/2 ,x=-(3+2sqrt(6))/2

Find the zeros of 4x^2+12x-15=0

The zeros of 4x^2+12x-15=0 are x=(-3+2sqrt(6))/2 ,x=-(3+2sqrt(6))/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

4 x^{2} + 12 x - 15 = 0

x = \frac{- 3 + 2 6}{2}, x = - \frac{3 + 2 6}{2}

Solution steps

4 x^{2} + 12 x - 15 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 12 \pm 1 2 ^{2} - 4 \cdot 4 ( - 15 )}{2 \cdot 4}

1 2^{2} - 4 \cdot 4 (- 15) = 86

x_{1, 2} = \frac{- 12 \pm 8 6}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- 12 + 8 6}{2 \cdot 4}, x_{2} = \frac{- 12 - 8 6}{2 \cdot 4}

x = \frac{- 12 + 8 6}{2 \cdot 4} : \frac{- 3 + 2 6}{2}

x = \frac{- 12 - 8 6}{2 \cdot 4} : - \frac{3 + 2 6}{2}

The solutions to the quadratic equation are:

x = \frac{- 3 + 2 6}{2}, x = - \frac{3 + 2 6}{2}

c^{3} - 8

7 x = \frac{14 + 7 x}{2}

y - 4.6 = 12

2 x^{2} + 11 x + 5 = 0

12 x^{2} - 2 x - 3 = 0

What are the solutions to the equation 4x^2+12x-15=0 ?
The solutions to the equation 4x^2+12x-15=0 are x=(-3+2sqrt(6))/2 ,x=-(3+2sqrt(6))/2
Find the zeros of 4x^2+12x-15=0
The zeros of 4x^2+12x-15=0 are x=(-3+2sqrt(6))/2 ,x=-(3+2sqrt(6))/2