What are the solutions to the equation x^2-9x+19=x-5 ?

The solutions to the equation x^2-9x+19=x-5 are x=6,x=4

Find the zeros of x^2-9x+19=x-5

The zeros of x^2-9x+19=x-5 are x=6,x=4

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x^{2} - 9 x + 19 = x - 5

x = 6, x = 4

Solution steps

x^{2} - 9 x + 19 = x - 5

Move

5

to the left side

x^{2} - 9 x + 24 = x

Move

x

to the left side

x^{2} - 10 x + 24 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 10 ) \pm ( - 10 ) ^{2} - 4 \cdot 1 \cdot 24}{2 \cdot 1}

(- 10)^{2} - 4 \cdot 1 \cdot 24 = 2

x_{1, 2} = \frac{- ( - 10 ) \pm 2}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 10 ) + 2}{2 \cdot 1}, x_{2} = \frac{- ( - 10 ) - 2}{2 \cdot 1}

x = \frac{- ( - 10 ) + 2}{2 \cdot 1} : 6

x = \frac{- ( - 10 ) - 2}{2 \cdot 1} : 4

The solutions to the quadratic equation are:

x = 6, x = 4

x^{2} - 32 x + 256

3 - 23 x = - 14 (x - 21) - 15 (x + 33)

\frac{2}{3} \cdot 1

\frac{2}{x + 5} > \frac{1}{x - 4}

What are the solutions to the equation x^2-9x+19=x-5 ?
The solutions to the equation x^2-9x+19=x-5 are x=6,x=4
Find the zeros of x^2-9x+19=x-5
The zeros of x^2-9x+19=x-5 are x=6,x=4