# Solve the following quadratic equations by factorization method: x^{2}+x-(a+1)(a+2)=0

Solve the following quadratic equations by factorization method:
${x}^{2}+x-\left(a+1\right)\left(a+2\right)=0$
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Step 1
${x}^{2}+x-\left(a+1\right)\left(a+2\right)=0$
Step 2
So
$⇒{x}^{2}+\left(a+2\right)x-\left(a+1\right)x-\left(a+1\right)\left(a+2\right)=0$
$⇒x\left(x+\left(a+2\right)\right)-\left(a+1\right)\left(x+\left(a+2\right)\right)=0$
$⇒\left(x+\left(a+2\right)\right)\left(x-\left(a+1\right)\right)=0$
$⇒$ x=-(a+2) or x=(a+1)
Hence x=-(a+2)
or x=(a+1)

Deufemiak7