(3/4)g-9=4(g+1)

Solution for (3/4)g-9=4(g+1) equation:

(3/4)g-9=4(g+1)

We move all terms to the left:

(3/4)g-9-(4(g+1))=0


Domain of the equation: 4)g!=0

g!=0/1

g!=0

g∈R


We add all the numbers together, and all the variables

(+3/4)g-(4(g+1))-9=0

We multiply parentheses

3g^2-(4(g+1))-9=0


We calculate terms in parentheses: -(4(g+1)), so:

4(g+1)

We multiply parentheses

4g+4

Back to the equation:

-(4g+4)


We get rid of parentheses

3g^2-4g-4-9=0

We add all the numbers together, and all the variables

3g^2-4g-13=0

a = 3; b = -4; c = -13;
Δ = b2-4ac
Δ = -42-4·3·(-13)
Δ = 172
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:

$\sqrt{\Delta}=\sqrt{172}=\sqrt{4*43}=\sqrt{4}*\sqrt{43}=2\sqrt{43}$
$g_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-2\sqrt{43}}{2*3}=\frac{4-2\sqrt{43}}{6}
$
$g_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+2\sqrt{43}}{2*3}=\frac{4+2\sqrt{43}}{6}
$