What are the solutions to the equation 2x^2-3x-3=0 ?

The solutions to the equation 2x^2-3x-3=0 are x=(3+sqrt(33))/4 ,x=(3-sqrt(33))/4

Find the zeros of 2x^2-3x-3=0

The zeros of 2x^2-3x-3=0 are x=(3+sqrt(33))/4 ,x=(3-sqrt(33))/4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

x = \frac{3 + 33}{4}, x = \frac{3 - 33}{4}

Decimal Notation

x = 2.18614 \dots, x = - 0.68614 \dots

Solution steps

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

Solve with the quadratic formula

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

(- 3)^{2} - 4 \cdot 2 (- 3) = 33

x_{1, 2} = \frac{- ( - 3 ) \pm 33}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 3 ) + 33}{2 \cdot 2}, x_{2} = \frac{- ( - 3 ) - 33}{2 \cdot 2}

x = \frac{- ( - 3 ) + 33}{2 \cdot 2} : \frac{3 + 33}{4}

x = \frac{- ( - 3 ) - 33}{2 \cdot 2} : \frac{3 - 33}{4}

The solutions to the quadratic equation are:

x = \frac{3 + 33}{4}, x = \frac{3 - 33}{4}

7 x (w - 4) + 5 (w - 4)

6 x^{2} - 5 x - 6 > 0

6 x - 3 = 2 x - 3 - 2 x

\frac{4}{5} (\frac{3}{4} - \frac{6}{7})

\frac{2}{3 - 1}

What are the solutions to the equation 2x^2-3x-3=0 ?
The solutions to the equation 2x^2-3x-3=0 are x=(3+sqrt(33))/4 ,x=(3-sqrt(33))/4
Find the zeros of 2x^2-3x-3=0
The zeros of 2x^2-3x-3=0 are x=(3+sqrt(33))/4 ,x=(3-sqrt(33))/4