Question
Solve the following quadratic equation by factorization method
$3\left(p^2-6\right)=p(p+5)$
$3\left(p^2-6\right)=p(p+5)$
✓
Answer
$3 p^2-18=p^2+5 p$
$2 p^2-5 p-18=0$
$2 p^2-9 p+4 p-18=0$
$p(2 p-9)+2(2 p-9)=0$
$(2 p-9)(p+2)=0$
$2 p-9=0 \text { or } p+2=0$
The roots are $p=\frac{9}{2},-2$
$2 p^2-5 p-18=0$
$2 p^2-9 p+4 p-18=0$
$p(2 p-9)+2(2 p-9)=0$
$(2 p-9)(p+2)=0$
$2 p-9=0 \text { or } p+2=0$
The roots are $p=\frac{9}{2},-2$
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