What are the solutions to the equation -3x+x^2-4=-8+x ?

The solutions to the equation -3x+x^2-4=-8+x are x=2

Find the zeros of -3x+x^2-4=-8+x

The zeros of -3x+x^2-4=-8+x are x=2

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- 3 x + x^{2} - 4 = - 8 + x

x = 2

Solution steps

- 3 x + x^{2} - 4 = - 8 + x

Move

x

to the left side

x^{2} - 4 x - 4 = - 8

Move

8

to the left side

x^{2} - 4 x + 4 = 0

Solve with the quadratic formula

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

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

x_{1, 2} = \frac{- ( - 4 ) \pm 0}{2 \cdot 1}

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

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

x = 2

The solution to the quadratic equation is:

x = 2

6 x (x - 1) = 63 - 19 x

6 x^{2} - x - 15 = 0

(x - 9) (x + 7) = 1

4 (x^{2} - x) = x^{2} - 3

4 (x^{2} - x) = x^{2} - 2

What are the solutions to the equation -3x+x^2-4=-8+x ?
The solutions to the equation -3x+x^2-4=-8+x are x=2
Find the zeros of -3x+x^2-4=-8+x
The zeros of -3x+x^2-4=-8+x are x=2