Solve x_1^2-2-(x-08)=-2(09+x)

$$x _ { 1 } ^ { 2 } - 2 - ( x - 08 ) = - 2 ( 09 + x )$$

Solve for x

$$x=2-x_{1}^{2}$$

Solution Steps

Multiply 0 and 8 to get 0.

$$x_{1}^{2}-2-\left(x-0\right)=-2\left(0\times 9+x\right)$$

Multiply 0 and 9 to get 0.

$$x_{1}^{2}-2-\left(x-0\right)=-2x$$

Add 2x to both sides.

$$x_{1}^{2}-2-\left(x-0\right)+2x=0$$

Reorder the terms.

$$x_{1}^{2}-2-x+2x=0$$

Combine -x and 2x to get x.

$$x_{1}^{2}-2+x=0$$

Subtract x_{1}^{2} from both sides. Anything subtracted from zero gives its negation.

$$-2+x=-x_{1}^{2}$$

Add 2 to both sides.

$$x=-x_{1}^{2}+2$$

Solve for x_1 (complex solution)

$$x_{1}=-\sqrt{2-x}$$

$$x_{1}=\sqrt{2-x}$$

Solve for x_1

$$x_{1}=\sqrt{2-x}$$

$$x_{1}=-\sqrt{2-x}\text{, }x\leq 2$$