Answer
$x=-\dfrac{3}{4}\pm\dfrac{\sqrt{17}}{4}$ satisfies the given conditions.
Work Step by Step
$y_{1}=-x^{2}+4x-2$ $,$ $y_{2}=-3x^{2}+x-1$ and $y_{1}-y_{2}=0$
If $y_{1}-y_{2}$ is equal to $0$, then $(-x^{2}+4x-2)-(-3x^{2}+x-1)$ is also equal to $0$.
$(-x^{2}+4x-2)-(-3x^{2}+x-1)=0$
Simplify:
$-x^{2}+4x-2+3x^{2}-x+1=0$
$2x^{2}+3x-1=0$
Use the quadratic formula to solve this equation. The formula is $x=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$
In this case, $a=2$, $b=3$ and $c=-1$
Substitute the known values into the formula and evaluate:
$x=\dfrac{-3\pm\sqrt{3^{2}-4(2)(-1)}}{2(2)}=\dfrac{-3\pm\sqrt{9+8}}{4}=\dfrac{-3\pm\sqrt{17}}{4}=...$
$...=-\dfrac{3}{4}\pm\dfrac{\sqrt{17}}{4}$
$x=-\dfrac{3}{4}\pm\dfrac{\sqrt{17}}{4}$ satisfies the given conditions.