Question 873564
Simplify y2 equation and put into general form, decreasing powers of x; and then equate corresponding coefficients.  You will have three equations in the unknowns, A, B, and C.
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Skipping the steps here, but done separately on paper, the y2 expression becomes
{{{Cx^2+Ax+Bx-4Cx+10A-5B-9C}}}
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Are you still stuck?
"yes".


Simplifying the y2 equation's right side:
{{{Ax+10A+Bx-5B+Cx^2-4Cx-9C}}}
{{{Cx^2+Ax+Bx-4Cx+10A-5B-9C}}}
{{{Cx^2+(A+B-4C)x+(10A-5B-9C)}}}


Now as given y1 equation to be equal to y2 eqation,
{{{-4x^2+9x-10=Cx^2+(A+B-4C)x+(10A-5B-9C)}}}
The coefficients in corresponding positions are equal.  This gives a system,
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-4=C
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9=A+B-4C
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-10=10A-5B-9C


Solve that system for A, B, and C.