What are the solutions to the equation 2(2x-5)-x^2+17x=16*4 ?

The solutions to the equation 2(2x-5)-x^2+17x=16*4 are x=(21-sqrt(145))/2 ,x=(21+sqrt(145))/2

Find the zeros of 2(2x-5)-x^2+17x=16*4

The zeros of 2(2x-5)-x^2+17x=16*4 are x=(21-sqrt(145))/2 ,x=(21+sqrt(145))/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

2 (2 x - 5) - x^{2} + 17 x = 16 \cdot 4

x = \frac{21 - 145}{2}, x = \frac{21 + 145}{2}

Decimal Notation

x = 4.47920 \dots, x = 16.52079 \dots

Solution steps

2 (2 x - 5) - x^{2} + 17 x = 16 \cdot 4

Expand

2 (2 x - 5) - x^{2} + 17 x : - x^{2} + 21 x - 10

Expand

16 \cdot 4 : 64

- x^{2} + 21 x - 10 = 64

Move

64

to the left side

- x^{2} + 21 x - 74 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 21 \pm 2 1 ^{2} - 4 ( - 1 ) ( - 74 )}{2 ( - 1 )}

2 1^{2} - 4 (- 1) (- 74) = 145

x_{1, 2} = \frac{- 21 \pm 145}{2 ( - 1 )}

Separate the solutions

x_{1} = \frac{- 21 + 145}{2 ( - 1 )}, x_{2} = \frac{- 21 - 145}{2 ( - 1 )}

x = \frac{- 21 + 145}{2 ( - 1 )} : \frac{21 - 145}{2}

x = \frac{- 21 - 145}{2 ( - 1 )} : \frac{21 + 145}{2}

The solutions to the quadratic equation are:

x = \frac{21 - 145}{2}, x = \frac{21 + 145}{2}

5 x (x + 1) = 15

- 3 (x - 2) - 4 x^{2} = 0

x^{2} - 8 x - 8 = 0

4 x^{2} + 4 x + 1 = 0

(2 x - 8) (2 x + 8) = 55

What are the solutions to the equation 2(2x-5)-x^2+17x=16*4 ?
The solutions to the equation 2(2x-5)-x^2+17x=16*4 are x=(21-sqrt(145))/2 ,x=(21+sqrt(145))/2
Find the zeros of 2(2x-5)-x^2+17x=16*4
The zeros of 2(2x-5)-x^2+17x=16*4 are x=(21-sqrt(145))/2 ,x=(21+sqrt(145))/2