What are the solutions to the equation x^2-5x+33=7x-3 ?

The solutions to the equation x^2-5x+33=7x-3 are x=6

Find the zeros of x^2-5x+33=7x-3

The zeros of x^2-5x+33=7x-3 are x=6

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x^{2} - 5 x + 33 = 7 x - 3

x = 6

Solution steps

x^{2} - 5 x + 33 = 7 x - 3

Move

3

to the left side

x^{2} - 5 x + 36 = 7 x

Move

7 x

to the left side

x^{2} - 12 x + 36 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 12 ) \pm ( - 12 ) ^{2} - 4 \cdot 1 \cdot 36}{2 \cdot 1}

(- 12)^{2} - 4 \cdot 1 \cdot 36 = 0

x_{1, 2} = \frac{- ( - 12 ) \pm 0}{2 \cdot 1}

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

\frac{- ( - 12 )}{2 \cdot 1} = 6

x = 6

The solution to the quadratic equation is:

x = 6

8^{x} = 5^{2 x - 1}

\frac{7 x - 4}{3} - \frac{5 - 8 x}{- 3} \geq \frac{4 x - 7}{3}

x + 6 = 0

(- 2)^{2} - 2 (- 2)

\frac{6 x + 4}{3} = \frac{5 x - 5}{8}

What are the solutions to the equation x^2-5x+33=7x-3 ?
The solutions to the equation x^2-5x+33=7x-3 are x=6
Find the zeros of x^2-5x+33=7x-3
The zeros of x^2-5x+33=7x-3 are x=6