What are the solutions to the equation 0=x^2-16x+48 ?

The solutions to the equation 0=x^2-16x+48 are x=12,x=4

Find the zeros of 0=x^2-16x+48

The zeros of 0=x^2-16x+48 are x=12,x=4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

0 = x^{2} - 16 x + 48

x = 12, x = 4

Solution steps

0 = x^{2} - 16 x + 48

Switch sides

x^{2} - 16 x + 48 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 16 ) \pm ( - 16 ) ^{2} - 4 \cdot 1 \cdot 48}{2 \cdot 1}

(- 16)^{2} - 4 \cdot 1 \cdot 48 = 8

x_{1, 2} = \frac{- ( - 16 ) \pm 8}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 16 ) + 8}{2 \cdot 1}, x_{2} = \frac{- ( - 16 ) - 8}{2 \cdot 1}

x = \frac{- ( - 16 ) + 8}{2 \cdot 1} : 12

x = \frac{- ( - 16 ) - 8}{2 \cdot 1} : 4

The solutions to the quadratic equation are:

x = 12, x = 4

(\frac{85}{4})^{\frac{3}{2}}

49 x^{3} - x y^{2}

4 w \geq 8

2^{2} - 6 (2) + 9

3 t^{(2)} + 9 t + 6 = 0

What are the solutions to the equation 0=x^2-16x+48 ?
The solutions to the equation 0=x^2-16x+48 are x=12,x=4
Find the zeros of 0=x^2-16x+48
The zeros of 0=x^2-16x+48 are x=12,x=4