What are the solutions to the equation 2x^2-3x=7 ?

The solutions to the equation 2x^2-3x=7 are x=(3+sqrt(65))/4 ,x=(3-sqrt(65))/4

Find the zeros of 2x^2-3x=7

The zeros of 2x^2-3x=7 are x=(3+sqrt(65))/4 ,x=(3-sqrt(65))/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 = 7

x = \frac{3 + 65}{4}, x = \frac{3 - 65}{4}

Decimal Notation

x = 2.76556 \dots, x = - 1.26556 \dots

Solution steps

2 x^{2} - 3 x = 7

Move

7

to the left side

2 x^{2} - 3 x - 7 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 2 ( - 7 )}{2 \cdot 2}

(- 3)^{2} - 4 \cdot 2 (- 7) = 65

x_{1, 2} = \frac{- ( - 3 ) \pm 65}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 3 ) + 65}{2 \cdot 2}, x_{2} = \frac{- ( - 3 ) - 65}{2 \cdot 2}

x = \frac{- ( - 3 ) + 65}{2 \cdot 2} : \frac{3 + 65}{4}

x = \frac{- ( - 3 ) - 65}{2 \cdot 2} : \frac{3 - 65}{4}

The solutions to the quadratic equation are:

x = \frac{3 + 65}{4}, x = \frac{3 - 65}{4}

y, 3 x^{2} - y^{2} = 5 x^{3} y^{2} - 4

x, 3 x^{2} - 8 x + 4 = 0

x, y = c_{1} x^{2} + c_{2} x + c_{3}

0 = 5 x^{2} - 20

64 x^{2} + 9 = 0

What are the solutions to the equation 2x^2-3x=7 ?
The solutions to the equation 2x^2-3x=7 are x=(3+sqrt(65))/4 ,x=(3-sqrt(65))/4
Find the zeros of 2x^2-3x=7
The zeros of 2x^2-3x=7 are x=(3+sqrt(65))/4 ,x=(3-sqrt(65))/4