What are the solutions to the equation r^2-10r+24=0 ?

The solutions to the equation r^2-10r+24=0 are r=6,r=4

Find the zeros of r^2-10r+24=0

The zeros of r^2-10r+24=0 are r=6,r=4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

r^{2} - 10 r + 24 = 0

r = 6, r = 4

Solution steps

r^{2} - 10 r + 24 = 0

Solve with the quadratic formula

r_{1, 2} = \frac{- ( - 10 ) \pm ( - 10 ) ^{2} - 4 \cdot 1 \cdot 24}{2 \cdot 1}

(- 10)^{2} - 4 \cdot 1 \cdot 24 = 2

r_{1, 2} = \frac{- ( - 10 ) \pm 2}{2 \cdot 1}

Separate the solutions

r_{1} = \frac{- ( - 10 ) + 2}{2 \cdot 1}, r_{2} = \frac{- ( - 10 ) - 2}{2 \cdot 1}

r = \frac{- ( - 10 ) + 2}{2 \cdot 1} : 6

r = \frac{- ( - 10 ) - 2}{2 \cdot 1} : 4

The solutions to the quadratic equation are:

r = 6, r = 4

\frac{i}{( i - 1 ) ( i - 2 ) ( i - 3 )}

3 \cdot (4 - x) \geq 18 x + 5

64 p^{2} = 9

2 x^{2} - 5 x - 375

4 i \cdot (- 7 + i)

What are the solutions to the equation r^2-10r+24=0 ?
The solutions to the equation r^2-10r+24=0 are r=6,r=4
Find the zeros of r^2-10r+24=0
The zeros of r^2-10r+24=0 are r=6,r=4