What are the solutions to the equation -2x^2-16x-28=0 ?

The solutions to the equation -2x^2-16x-28=0 are x=-4-sqrt(2),x=-4+sqrt(2)

Find the zeros of -2x^2-16x-28=0

The zeros of -2x^2-16x-28=0 are x=-4-sqrt(2),x=-4+sqrt(2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

- 2 x^{2} - 16 x - 28 = 0

x = - 4 - 2, x = - 4 + 2

Decimal Notation

x = - 5.41421 \dots, x = - 2.58578 \dots

Solution steps

- 2 x^{2} - 16 x - 28 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 16 ) \pm ( - 16 ) ^{2} - 4 ( - 2 ) ( - 28 )}{2 ( - 2 )}

(- 16)^{2} - 4 (- 2) (- 28) = 42

x_{1, 2} = \frac{- ( - 16 ) \pm 4 2}{2 ( - 2 )}

Separate the solutions

x_{1} = \frac{- ( - 16 ) + 4 2}{2 ( - 2 )}, x_{2} = \frac{- ( - 16 ) - 4 2}{2 ( - 2 )}

x = \frac{- ( - 16 ) + 4 2}{2 ( - 2 )} : - 4 - 2

x = \frac{- ( - 16 ) - 4 2}{2 ( - 2 )} : - 4 + 2

The solutions to the quadratic equation are:

x = - 4 - 2, x = - 4 + 2

\frac{13 !}{( 13 - 3 ) !}

3 ∣ x - 5 ∣ > 9

x + 7 = 4

x + 1 \geq 0

23 \cdot 23

What are the solutions to the equation -2x^2-16x-28=0 ?
The solutions to the equation -2x^2-16x-28=0 are x=-4-sqrt(2),x=-4+sqrt(2)
Find the zeros of -2x^2-16x-28=0
The zeros of -2x^2-16x-28=0 are x=-4-sqrt(2),x=-4+sqrt(2)