What are the two solutions of x2 – 2x – 4 = –3x + 9?
the y-coordinates of the y-intercepts of the graphs of y = x2 – 2x – 4 and y = –3x + 9
the x-coordinates of the x-intercepts of the graphs of y = x2 – 2x – 4 and y = –3x + 9
the y-coordinates of the intersection points of the graphs of y = x2 – 2x – 4 and y = –3x + 9
the x-coordinates of the intersection points of the graphs of y = x2 – 2x – 4 and y = –3x + 9

"It is a polynomial equation with degree two."
"The general form of quadratic equation is [tex]ax^2+bx+c=0[/tex] , where a, b, c are real numbers with a ≠ 0."

For given question,

We have been given an quadratic equation [tex]x^2 - 2x - 4 = -3x + 9[/tex]

We simplify above quadratic equation.

[tex]\Rightarrow x^2 - 2x - 4 = -3x + 9\\\Rightarrow x^2 - 2x+3x - 4-9 = -3x + 9+3x-9\\\Rightarrow x^2+x-13=0\\\Rightarrow (x-3.14)(x+4.14)=0\\\Rightarrow x=3.14~~or~~x=-4.14[/tex]

Consider the graphs for y = x² - 2x - 4 and y = -3x + 9 as shown below.

Therefore, the solutions of x² - 2x - 4 = -3x + 9 are x = -4.14 and x = 3.14,

which is the x-coordinates of the intersection points of the graphs of y = x² - 2x - 4 and y = -3x + 9

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