A quadratic follows the format:
$f(x)=Ax^2+Bx+C$
Now we know $f(0)=0.15$, $f(0.5)=0.1$ and $f(1)=0.08$. Plugging these into our original formula, we get $3$ equations with $3$ variables. If you can solve this to obtain $A$, $B$ and $C$, you have found your quadratic equation. To solve them, you should try substitution.
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$f(0)=A \cdot 0^2+ B \cdot 0+C$
$0.15=C$
So now we can write $f(x)=Ax^2+Bx+0.15$.
$f(0.5)=\frac A4+\frac B2+0.15=0.1$
$A+2B=-0.2 \tag 1$
$f(1)=A+B+0.15=0.08$
$A+B=-0.07 \tag 2$
Subtracting $(2)$ from $(1)$:
$B=-0.13$
Substituting this into $(2)$,
$A-0.13=-0.07 \implies A=0.06$.
Therefore $f(x)=0.06x^2-0.13x+0.15$