3a^2+22a+7=(a+7)

Solution for 3a^2+22a+7=(a+7) equation:

3a^2+22a+7=(a+7)

We move all terms to the left:

3a^2+22a+7-((a+7))=0


We calculate terms in parentheses: -((a+7)), so:

(a+7)

We get rid of parentheses

a+7

Back to the equation:

-(a+7)


We get rid of parentheses

3a^2+22a-a-7+7=0

We add all the numbers together, and all the variables

3a^2+21a=0

a = 3; b = 21; c = 0;
Δ = b2-4ac
Δ = 212-4·3·0
Δ = 441
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{441}=21$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(21)-21}{2*3}=\frac{-42}{6}
=-7
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(21)+21}{2*3}=\frac{0}{6}
=0
$