What are the solutions to the equation x^2-22x+72=0 ?

The solutions to the equation x^2-22x+72=0 are x=18,x=4

Find the zeros of x^2-22x+72=0

The zeros of x^2-22x+72=0 are x=18,x=4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

x^{2} - 22 x + 72 = 0

x = 18, x = 4

Solution steps

x^{2} - 22 x + 72 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 22 ) \pm ( - 22 ) ^{2} - 4 \cdot 1 \cdot 72}{2 \cdot 1}

(- 22)^{2} - 4 \cdot 1 \cdot 72 = 14

x_{1, 2} = \frac{- ( - 22 ) \pm 14}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 22 ) + 14}{2 \cdot 1}, x_{2} = \frac{- ( - 22 ) - 14}{2 \cdot 1}

x = \frac{- ( - 22 ) + 14}{2 \cdot 1} : 18

x = \frac{- ( - 22 ) - 14}{2 \cdot 1} : 4

The solutions to the quadratic equation are:

x = 18, x = 4

0 = 3 - x^{2}

x, x^{2} - 2 k x + 3 k = 0

x^{2} - x - 12 = 0

7 a + 2 = - 2 a^{2}

x^{2} + 7 x + \frac{33}{4} = 0

What are the solutions to the equation x^2-22x+72=0 ?
The solutions to the equation x^2-22x+72=0 are x=18,x=4
Find the zeros of x^2-22x+72=0
The zeros of x^2-22x+72=0 are x=18,x=4