Find all points, if any, where y - 4x = 12 intersects 2 - y = 2(x + 2)2 There are no points of intersection (-3, 0) (3, 0) (-3, 0) and (3, 0)

Answer:(-3, 0)Step-by-step explanation:We need to solve to the system of equations:[tex]\left\{\begin{array}{ccc}y-4x=12\\2-y=2(x+2)^2\end{array}\right\\\\y-4x=12\qquad\text{add 4x to both sides}\\y=12+4x\qquad(1)\\\\\text{Substitute it the the second equation:}\\\\2-(12+4x)=2(x+2)^2\qquad\text{use}\ (a+b)^2=a^2+2ab+b^2\\\\2-12-4x=2(x^2+2(x)(2)+2^2)\\\\-10-4x=2(x^2+4x+4)\qquad\text{use distributive property}\\\\-10-4x=2x^2+8x+8\qquad\text{add 10 and 4x to both sides}\\\\0=2x^2+12x+18\qquad\text{divide both sides by 2}\\\\x^2+6x+9=0\\\\x^2+3x+3x+9=0\\\\x(x+3)+3(x+3)=0\\\\(x+3)(x+3)=0\iff x+3=0[/tex][tex]x=-3\\\\\text{Put the value of x to (1):}\\\\y=12+4(-3)\\\\y=12-12\\\\y=0[/tex]