Within 11year peter age will be half of the square of the age he was 13year ago find his current age and form quadratic equation

Let Peter's current age be $P$ years.
Age in 11 years: $P+11$.
Age 13 years ago: $P-13$.
According to the problem, Peter's age in 11 years will be half of the square of his age 13 years ago:

$$P+11 = \frac{1}{2}(P-13)^2$$

Multiply both sides by 2:

$$2(P+11) = (P-13)^2$$

Expand both sides:

$$2P + 22 = P^2 - 2(P)(13) + 13^2$$ $$2P + 22 = P^2 - 26P + 169$$

Move all terms to one side to form a standard quadratic equation $aP^2 + bP + c = 0$:

$$0 = P^2 - 26P - 2P + 169 - 22$$ $$0 = P^2 - 28P + 147$$

The quadratic equation is $\boxed{P^2 - 28P + 147 = 0}$.

We can factor the quadratic equation $P^2 - 28P + 147 = 0$. We need two numbers that multiply to 147 and add to -28. These numbers are -7 and -21.

$$(P-7)(P-21) = 0$$

This gives two possible values for $P$:

$$P-7 = 0 \Rightarrow P = 7$$ $$P-21 = 0 \Rightarrow P = 21$$

If $P=7$, Peter's age 13 years ago would be $7-13 = -6$. Age cannot be negative, so $P=7$ is not a valid solution.
If $P=21$:
Age in 11 years: $21+11 = 32$.
Age 13 years ago: $21-13 = 8$.
Half of the square of the age 13 years ago: $\frac{1}{2}(8)^2 = \frac{1}{2}(64) = 32$.
This matches the condition.

Peter's current age is $\boxed{21 \text{ years}}$.