What are the solutions to the equation (21-r)(28-r)-294=0 ?

The solutions to the equation (21-r)(28-r)-294=0 are r=42,r=7

Find the zeros of (21-r)(28-r)-294=0

The zeros of (21-r)(28-r)-294=0 are r=42,r=7

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(21 - r) (28 - r) - 294 = 0

r = 42, r = 7

Solution steps

(21 - r) (28 - r) - 294 = 0

Expand

(21 - r) (28 - r) - 294 : r^{2} - 49 r + 294

r^{2} - 49 r + 294 = 0

Solve with the quadratic formula

r_{1, 2} = \frac{- ( - 49 ) \pm ( - 49 ) ^{2} - 4 \cdot 1 \cdot 294}{2 \cdot 1}

(- 49)^{2} - 4 \cdot 1 \cdot 294 = 35

r_{1, 2} = \frac{- ( - 49 ) \pm 35}{2 \cdot 1}

Separate the solutions

r_{1} = \frac{- ( - 49 ) + 35}{2 \cdot 1}, r_{2} = \frac{- ( - 49 ) - 35}{2 \cdot 1}

r = \frac{- ( - 49 ) + 35}{2 \cdot 1} : 42

r = \frac{- ( - 49 ) - 35}{2 \cdot 1} : 7

The solutions to the quadratic equation are:

r = 42, r = 7

- 10 p + 9 p = 12

4 (1 - 3 x) = 52

- 2 - 1

5 + 4 [1 + 2 (2 x - 4)] = 6 (x + 5)

4 + \frac{r}{6} \leq 5

What are the solutions to the equation (21-r)(28-r)-294=0 ?
The solutions to the equation (21-r)(28-r)-294=0 are r=42,r=7
Find the zeros of (21-r)(28-r)-294=0
The zeros of (21-r)(28-r)-294=0 are r=42,r=7