Given:
[tex]\begin{gathered} y=3x^2\text{ + 13x -50} \\ y\text{ = 13x }-\text{ 2} \end{gathered}[/tex]Subtracting equation 2 from 1:
[tex]\begin{gathered} y-y\text{ = }3x^2\text{ + 13x - 50 -(13x - 2)} \\ 0=3x^2\text{ + 13x - 50 - 13x + 2} \\ 3x^2\text{ -48 = 0} \end{gathered}[/tex]Solving for x:
[tex]\begin{gathered} 3x^2\text{ - 48 = 0} \\ 3x^2\text{ = 48} \\ \text{Divide both sides by 3} \\ x^2\text{ = }\frac{48}{3} \\ x^2\text{ = 16} \\ \text{Square root both sides} \\ x\text{ = }\sqrt[]{16} \\ x\text{ = }\pm\text{ 4} \end{gathered}[/tex]Substituting the value of x into equation 2:
[tex]\begin{gathered} y\text{ = 13x - 2} \\ y\text{ = 13(}\pm4)\text{ - 2} \\ y\text{ = 52 - 2 } \\ =\text{ 50} \\ or\text{ } \\ y\text{ = -52 - 2} \\ =\text{ -54} \end{gathered}[/tex]Hence, the solution to the system of equations is:
(4, 50) and (-4 , -54)