What are the solutions to the equation a^2+12=7a ?

The solutions to the equation a^2+12=7a are a=4,a=3

Find the zeros of a^2+12=7a

The zeros of a^2+12=7a are a=4,a=3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

a^{2} + 12 = 7 a

a = 4, a = 3

Solution steps

a^{2} + 12 = 7 a

Move

7 a

to the left side

a^{2} + 12 - 7 a = 0

Write in the standard form

a x^{2} + b x + c = 0

a^{2} - 7 a + 12 = 0

Solve with the quadratic formula

a_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 \cdot 1 \cdot 12}{2 \cdot 1}

(- 7)^{2} - 4 \cdot 1 \cdot 12 = 1

a_{1, 2} = \frac{- ( - 7 ) \pm 1}{2 \cdot 1}

Separate the solutions

a_{1} = \frac{- ( - 7 ) + 1}{2 \cdot 1}, a_{2} = \frac{- ( - 7 ) - 1}{2 \cdot 1}

a = \frac{- ( - 7 ) + 1}{2 \cdot 1} : 4

a = \frac{- ( - 7 ) - 1}{2 \cdot 1} : 3

The solutions to the quadratic equation are:

a = 4, a = 3

h^{2} = 169

0 = - x^{2} + 2 x

(- 128)^{\frac{3}{7}}

- 9 \cdot (\frac{0}{- 3})

- 2 (6 + x) = 18 - 3 x

What are the solutions to the equation a^2+12=7a ?
The solutions to the equation a^2+12=7a are a=4,a=3
Find the zeros of a^2+12=7a
The zeros of a^2+12=7a are a=4,a=3