11g(g-1)+1=3(g+2)

Solution for 11g(g-1)+1=3(g+2) equation:

11g(g-1)+1=3(g+2)

We move all terms to the left:

11g(g-1)+1-(3(g+2))=0

We multiply parentheses

11g^2-11g-(3(g+2))+1=0


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

3(g+2)

We multiply parentheses

3g+6

Back to the equation:

-(3g+6)


We get rid of parentheses

11g^2-11g-3g-6+1=0

We add all the numbers together, and all the variables

11g^2-14g-5=0

a = 11; b = -14; c = -5;
Δ = b2-4ac
Δ = -142-4·11·(-5)
Δ = 416
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$g_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$g_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{416}=\sqrt{16*26}=\sqrt{16}*\sqrt{26}=4\sqrt{26}$
$g_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-4\sqrt{26}}{2*11}=\frac{14-4\sqrt{26}}{22}
$
$g_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+4\sqrt{26}}{2*11}=\frac{14+4\sqrt{26}}{22}
$