Answer
No real solution
Work Step by Step
$y=3x^2-4x-10$ ................... eq(1)
$y=-2x^2+6x-18$ ..................... eq(2)
Comparing eq(1) and eq(2)
$3x^2-4x-10=-2x^2+6x-18$
$3x^2-4x-10+2x^2-6x+18=0$
$3x^2+2x^2-4x-6x-10+18=0$
$5x^2-10x+8=0$
Comparing it with standard quadratic equation
$ax^2+bx+c=0$
$a=5,b=-10,c=8$
Putting in the formula
$x=\frac{-b \pm \sqrt {b^2-4ac}}{2a}$
$x=\frac{10 \pm \sqrt {(-10)^2-4(5)(8)}}{2(5)}$
$x=\frac{10 \pm \sqrt {(-60)}}{10}$
Which is not real,so solution set is empty.
