4(1+3)=3a(2a+1)

Solution for 4(1+3)=3a(2a+1) equation:

4(1+3)=3a(2a+1)

We move all terms to the left:

4(1+3)-(3a(2a+1))=0

We add all the numbers together, and all the variables

-(3a(2a+1))+44=0


We calculate terms in parentheses: -(3a(2a+1)), so:

3a(2a+1)

We multiply parentheses

6a^2+3a

Back to the equation:

-(6a^2+3a)


We get rid of parentheses

-6a^2-3a+44=0

a = -6; b = -3; c = +44;
Δ = b2-4ac
Δ = -32-4·(-6)·44
Δ = 1065
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-\sqrt{1065}}{2*-6}=\frac{3-\sqrt{1065}}{-12}
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{1065}}{2*-6}=\frac{3+\sqrt{1065}}{-12}
$