Answer
One solution: $ x=-\frac{9}{5}$
Work Step by Step
Adjust as follows:
$\sqrt{x^2+3x+7}=\sqrt{x^2-3x+9}-2$
Square each side:
$(\sqrt{x^2+3x+7})^2=(\sqrt{x^2-3x+9}-2)^2$
$x^2+3x+7=x^2-3x+9-4\sqrt{x^2-3x+9}+4$
Add same term variables:
$x^2+3x+7-(x^2-3x+9)-4=-4\sqrt{x^2-3x+9}$
$6x-6=-4\sqrt{x^2-3x+9}$
Square each side again:
$(6x-6)^2=(-4\sqrt{x^2-3x+9})^2$
$36x^2-36x-36x+36=16(x^2-3x+9)$
Simplify:
$36x^2-72x+36=16x^2-48x+144$
$36x^2-72x+36-(16x^2-48x+144)=0$
$20x^2-24x-108=0$
$4(5x^2-6x-27)=0$
$5x^2-6x-27=0$
Factor:
$(5x+9)(x-3)=0$
And we get two solutions:
$5x_1+9=0$
$5x_1=-9$
$x_1=-\frac{9}{5}$
$x_2-3=0$
$x_2=3$
We insert them into the original equation to find extraneous solutions:
$\sqrt{x_1^2+3x_1+7}-\sqrt{x_1^2-3x_1+9}+2=0$
$\sqrt{(-\frac{9}{5})^2+3(-\frac{9}{5})+7}-\sqrt{(-\frac{9}{5})^2-3(-\frac{9}{5})+9}+2=0$
$\sqrt{\frac{81}{25}-\frac{27}{5}+7}-\sqrt{\frac{81}{25}+\frac{27}{5}+9}+2=0$
$\sqrt{\frac{81}{25}-\frac{27(5)}{5(5)}+\frac{7(25)}{25}}-\sqrt{\frac{81}{25}+\frac{27(5)}{5(5)}+\frac{9(25)}{25}}+2=0$
$\sqrt{\frac{81}{25}-\frac{135}{25}+\frac{175}{25}}-\sqrt{\frac{81}{25}+\frac{135}{25}+\frac{225}{25}}+2=0$
$\sqrt{\frac{121}{25}}-\sqrt{\frac{441}{25}}+2=0$
$\frac{11}{5}-\frac{21}{5}+2=0$
$-\frac{10}{5}+2=0$
$-2+2=0$
$0=0\checkmark$
$\sqrt{x_2^2+3x_2+7}-\sqrt{x_2^2-3x_2+9}+2=0$
$\sqrt{(3^2+3(3)+7}-\sqrt{3^2-3(3)+9}+2=0$
$\sqrt{(9+9+7}-\sqrt{9-9+9}+2=0$
$\sqrt{25}-\sqrt{9}+2=0$
$5-3+2=0$
$4\ne0$
