$$(x+4)^{2}+(x-4)^{2}=2x(x-5)+8$$
$x = -\frac{12}{5} = -2\frac{2}{5} = -2.4$
$$x^{2}+8x+16+\left(x-4\right)^{2}=2x\left(x-5\right)+8$$
$$x^{2}+8x+16+x^{2}-8x+16=2x\left(x-5\right)+8$$
$$2x^{2}+8x+16-8x+16=2x\left(x-5\right)+8$$
$$2x^{2}+16+16=2x\left(x-5\right)+8$$
$$2x^{2}+32=2x\left(x-5\right)+8$$
$$2x^{2}+32=2x^{2}-10x+8$$
$$2x^{2}+32-2x^{2}=-10x+8$$
$$32=-10x+8$$
$$-10x+8=32$$
$$-10x=32-8$$
$$-10x=24$$
$$x=\frac{24}{-10}$$
$$x=-\frac{12}{5}$$
Show Solution
Hide Solution
