How to determine $x$ in $x^2+3x+7=4$? I'm helping my child with his homework. One of the problems is this: Determine the solution set of the equation
$x^2+3x+7=4$
Here is my attempt to determine $x_1$ and $x_2$:
$a^2+2ab+b^2=(a+b)^2$
$2ab=3x$
$2xb=3x$
$2b=3$
$b=\frac{3}{2}=1.5$
$b^2=2.25$
$x^2+3x+b^2-b^2+7=4$
$x^2+3x+2.25-2.25+7=4$
$(x^2+3x+2.25)-2.25+7=4$
$(x^2+3x+2.25)+7=4+2.25$
$(x^2+3x+2.25)+7=6.25$
$(x^2+3x+2.25)=6.25-7$
$(x^2+3x+2.25)=-0.75$
$(x+1.5)^2=-0.75$
In other similar equations we would take the square root of both sides. But I cannot take a square root of a negative number ($-0.75$).
How can I solve it, i. e. find out the values of $x$?

This is actually much simpler if you subtract the 4 to the left hand side first:
$$x^2 + 3x + 3 = 0$$
Then the solution set must satisfy the quadratic formula: 
$$x = \frac{-b\pm \sqrt{b^2-4ac}}{2a}$$
where we have the quadratic $ax^2 + bx + c = 0$. In this case, 
$$a=1, b=3, c = 3$$
So then $x$ must be $$x = \frac{-3 \pm \sqrt{3^2 - 4(1)(3)}}{2a}$$
However, we notice here that inside the square root, we have a negative number $(-3)$, which indicates that the quadratic has no solution set in the real numbers. 
If we are allowed to use imaginary numbers, then the solution would be 
$$x = \frac{-3\pm \sqrt{-3}}{2} = \frac{-3\pm i\sqrt{3}}{2}$$
