If sinθ and cosθ are the two roots of the quadratic equation $a{x}^2+bx+c=0$, then a, $b$ and $c$ satisfy:

The roots add to −$b$/a and multiply to $c$/a. Squaring the sum: $(sin\theta +cos\theta {)}^2=1+2$sinθ·cosθ, so $b^2/a^2=1+2c/a$, and multiplying through by $a^2$ gives $b^2=a^2+2ac$. Concrete case θ = 45°, where both roots are 1/√2: the equation is $x^2$ − √$2x+1/2=0$, and indeed $2=1+2(1/2)$. Losing the sign on the product gives $b^2=a^2-2ac$, dropping the factor 2 gives $b^2=a^2+ac$, and the relation cannot be rearranged into $a^2=b^2+2ac$.