What are the solutions to the equation 7x^2-3x+8=9x^2+15 ?

The solutions to the equation 7x^2-3x+8=9x^2+15 are x=-3/4-i(sqrt(47))/4 ,x=-3/4+i(sqrt(47))/4

Find the zeros of 7x^2-3x+8=9x^2+15

The zeros of 7x^2-3x+8=9x^2+15 are x=-3/4-i(sqrt(47))/4 ,x=-3/4+i(sqrt(47))/4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

7 x^{2} - 3 x + 8 = 9 x^{2} + 15

x = - \frac{3}{4} - i \frac{47}{4}, x = - \frac{3}{4} + i \frac{47}{4}

Solution steps

7 x^{2} - 3 x + 8 = 9 x^{2} + 15

Move

15

to the left side

7 x^{2} - 3 x - 7 = 9 x^{2}

Move

9 x^{2}

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 ( - 2 ) ( - 7 )}{2 ( - 2 )}

Simplify

(- 3)^{2} - 4 (- 2) (- 7) : 47 i

x_{1, 2} = \frac{- ( - 3 ) \pm 47 i}{2 ( - 2 )}

Separate the solutions

x_{1} = \frac{- ( - 3 ) + 47 i}{2 ( - 2 )}, x_{2} = \frac{- ( - 3 ) - 47 i}{2 ( - 2 )}

x = \frac{- ( - 3 ) + 47 i}{2 ( - 2 )} : - \frac{3}{4} - i \frac{47}{4}

x = \frac{- ( - 3 ) - 47 i}{2 ( - 2 )} : - \frac{3}{4} + i \frac{47}{4}

The solutions to the quadratic equation are:

x = - \frac{3}{4} - i \frac{47}{4}, x = - \frac{3}{4} + i \frac{47}{4}

15 x + x^{2} = 0

5 x^{2} - 12 a x + 3 a^{2} = 0

(\frac{2}{3} y - 24) (y - 36) = 240

81 - x^{2} = - 9

(x + 3)^{2} + (x + 2) (x - 2) = x (x - 1) - 5

What are the solutions to the equation 7x^2-3x+8=9x^2+15 ?
The solutions to the equation 7x^2-3x+8=9x^2+15 are x=-3/4-i(sqrt(47))/4 ,x=-3/4+i(sqrt(47))/4
Find the zeros of 7x^2-3x+8=9x^2+15
The zeros of 7x^2-3x+8=9x^2+15 are x=-3/4-i(sqrt(47))/4 ,x=-3/4+i(sqrt(47))/4