2(2n+3)=(4n^2-6n+9)

Solution for 2(2n+3)=(4n^2-6n+9) equation:

2(2n+3)=(4n^2-6n+9)

We move all terms to the left:

2(2n+3)-((4n^2-6n+9))=0

We multiply parentheses

4n-((4n^2-6n+9))+6=0


We calculate terms in parentheses: -((4n^2-6n+9)), so:

(4n^2-6n+9)

We get rid of parentheses

4n^2-6n+9

Back to the equation:

-(4n^2-6n+9)


We get rid of parentheses

-4n^2+4n+6n-9+6=0

We add all the numbers together, and all the variables

-4n^2+10n-3=0

a = -4; b = 10; c = -3;
Δ = b2-4ac
Δ = 102-4·(-4)·(-3)
Δ = 52
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{52}=\sqrt{4*13}=\sqrt{4}*\sqrt{13}=2\sqrt{13}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{13}}{2*-4}=\frac{-10-2\sqrt{13}}{-8}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{13}}{2*-4}=\frac{-10+2\sqrt{13}}{-8}
$