122=(2a+2)(2a)

Solution for 122=(2a+2)(2a) equation:

122=(2a+2)(2a)

We move all terms to the left:

122-((2a+2)(2a))=0


We calculate terms in parentheses: -((2a+2)2a), so:

(2a+2)2a

We multiply parentheses

4a^2+4a

Back to the equation:

-(4a^2+4a)


We get rid of parentheses

-4a^2-4a+122=0

a = -4; b = -4; c = +122;
Δ = b2-4ac
Δ = -42-4·(-4)·122
Δ = 1968
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1968}=\sqrt{16*123}=\sqrt{16}*\sqrt{123}=4\sqrt{123}$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4\sqrt{123}}{2*-4}=\frac{4-4\sqrt{123}}{-8}
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4\sqrt{123}}{2*-4}=\frac{4+4\sqrt{123}}{-8}
$