What are the solutions to the equation solvefor x,x^2-12x+32=0 ?

The solutions to the equation solvefor x,x^2-12x+32=0 are x=8,x=4

Find the zeros of solvefor x,x^2-12x+32=0

The zeros of solvefor x,x^2-12x+32=0 are x=8,x=4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

x, x^{2} - 12 x + 32 = 0

x = 8, x = 4

Solution steps

x^{2} - 12 x + 32 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 12 ) \pm ( - 12 ) ^{2} - 4 \cdot 1 \cdot 32}{2 \cdot 1}

(- 12)^{2} - 4 \cdot 1 \cdot 32 = 4

x_{1, 2} = \frac{- ( - 12 ) \pm 4}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 12 ) + 4}{2 \cdot 1}, x_{2} = \frac{- ( - 12 ) - 4}{2 \cdot 1}

x = \frac{- ( - 12 ) + 4}{2 \cdot 1} : 8

x = \frac{- ( - 12 ) - 4}{2 \cdot 1} : 4

The solutions to the quadratic equation are:

x = 8, x = 4

7 = - x^{2} + 16

5 x^{2} + 20 x + 8 = 0

m^{2} + 8 m + 6 = 0

8 x^{2} - 18 x + 7 = 0

5 x (x - 6) + 4 (x - 4) = - 7 x^{2} - 30

What are the solutions to the equation solvefor x,x^2-12x+32=0 ?
The solutions to the equation solvefor x,x^2-12x+32=0 are x=8,x=4
Find the zeros of solvefor x,x^2-12x+32=0
The zeros of solvefor x,x^2-12x+32=0 are x=8,x=4