## anonymous one year ago x

$$\bf \begin{cases} {\color{brown}{ y}}=-x^2+36\\ {\color{brown}{ y}}=2x+21 \end{cases}\qquad well\qquad \begin{array}{cccllll} {\color{brown}{ y}}&=&{\color{brown}{ y}} \\ \quad \\ -x^2+36&=&2x+21 \end{array}\qquad thus \\ \quad \\ -x^2+36=2x+21\implies 36=x^2+2x+21 \\ \quad \\ 0=x^2+2x+21-36\implies0=x^2+2x-15\qquad \\ \quad \\ notice\qquad \begin{array}{cccllll} x^2&+2x&-15\\ &\uparrow &\uparrow \\ &5-3&5\cdot -3 \end{array} \\ \quad \\ 0=(x+5)(x-3)$$ solve for "x" once you found "x", get "y" by substitution