Answer
$x=\dfrac{5}{3}\pm\dfrac{\sqrt{7}}{3}$ satisfies the given conditions.
Work Step by Step
$y_{1}=2x^{2}+5x-4$ $,$ $y_{2}=-x^{2}+15x-10$ and $y_{1}-y_{2}=0$
If $y_{1}-y_{2}$ is equal to $0$, then $(2x^{2}+5x-4)-(-x^{2}+15x-10)$ is also equal to $0$
$(2x^{2}+5x-4)-(-x^{2}+15x-10)=0$
Simplify:
$2x^{2}+5x-4+x^{2}-15x+10=0$
$3x^{2}-10x+6=0$
Use the quadratic formula to solve this equation. The formula is $x=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$
In this case, $a=3$, $b=-10$ and $c=6$
Substitute the known values into the formula and evaluate:
$x=\dfrac{-(-10)\pm\sqrt{(-10)^{2}-4(3)(6)}}{2(3)}=\dfrac{10\pm\sqrt{100-72}}{6}=...$
$...=\dfrac{10\pm\sqrt{28}}{6}=\dfrac{10\pm2\sqrt{7}}{6}=\dfrac{5}{3}\pm\dfrac{\sqrt{7}}{3}$
$x=\dfrac{5}{3}\pm\dfrac{\sqrt{7}}{3}$ satisfies the given conditions.
