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

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

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

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

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x^{2} + 12 = 7 x

x = 4, x = 3

Solution steps

x^{2} + 12 = 7 x

Move

7 x

to the left side

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

Write in the standard form

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

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

Solve with the quadratic formula

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

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

x = 4, x = 3

- 10 b + 7 = 17

4 - x = 2 x + 3 x - 5 x

x^{3} - 6 x^{2} + 12 x - 8

\frac{x ^{2} - x}{x ^{2} + 2 x} \leq 0

v^{2} = 16

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