Help please!!!Solve linear-quadratic system algebraically. Then what is the x coordinate of the solution? Show all work. y – 5 = (x – 2)^2x + 2y = 6

Answer:x = [tex]\frac{7+\sqrt{47}\times i }{4}[/tex]Step-by-step explanation:To solve quadratic systems,we always substitute one variable in terms if the other and then solve the equation.x + 2y = 6                                 ---------------(1)y - 5 = [tex](x-2)^{2}[/tex]         ---------------(2)y = [tex](x-2)^{2}[/tex] + 5         ---------------(3)Substitute (3) in (1) ,x + 2( [tex](x-2)^{2}[/tex] + 5 ) = 6[tex](a + b)^{2}[/tex] =[tex]a^{2} + 2ab + b^{2}[/tex]x + 2( [tex]x^{2} - 4x + 4[/tex] + 5 ) = 6[tex]2x^{2} - 7x + 12=0[/tex]      --------------(4)The roots of the quadratic equation [tex]ax^{2}  +bx+c[/tex] isx = [tex]\frac{(-b) + \sqrt{(-b)^{2}-4 \times ac }  }{2 \times a}[/tex]  -----------(5)According to equation (5),solution of (4) isx =  [tex]\frac{7 + \sqrt{(-7)^{2}-4 \times 24 }  }{2 \times 2}[/tex]x =  [tex]\frac{7+\sqrt{49-96}}{4}[/tex]x = [tex]\frac{7+\sqrt{47}\times i }{4}[/tex]