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

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

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

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

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

x = 2, x = - 4

Solution steps

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

Move

3 x

to the left side

x^{2} + 2 x - 8 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 1 \cdot ( - 8 )}{2 \cdot 1}

2^{2} - 4 \cdot 1 \cdot (- 8) = 6

x_{1, 2} = \frac{- 2 \pm 6}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 2 + 6}{2 \cdot 1}, x_{2} = \frac{- 2 - 6}{2 \cdot 1}

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

x = \frac{- 2 - 6}{2 \cdot 1} : - 4

The solutions to the quadratic equation are:

x = 2, x = - 4

16 d^{8} - 8 d^{4} + 1

24 x + 18 y + 39

z + 4 (2 z + 3) = 15

\frac{9}{2} (2 + x) \leq \frac{1}{3} (5 x + 14) \leq 11 - \frac{3 x}{2}

- 3 r^{2} - 2 r - 5 = 0

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