(3/4)g+16=34

Solution for (3/4)g+16=34 equation:

(3/4)g+16=34

We move all terms to the left:

(3/4)g+16-(34)=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+16-34=0

We add all the numbers together, and all the variables

(+3/4)g-18=0

We multiply parentheses

3g^2-18=0

a = 3; b = 0; c = -18;
Δ = b2-4ac
Δ = 02-4·3·(-18)
Δ = 216
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:

$\sqrt{\Delta}=\sqrt{216}=\sqrt{36*6}=\sqrt{36}*\sqrt{6}=6\sqrt{6}$
$g_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6\sqrt{6}}{2*3}=\frac{0-6\sqrt{6}}{6}
=-\frac{6\sqrt{6}}{6}
=-\sqrt{6}
$
$g_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6\sqrt{6}}{2*3}=\frac{0+6\sqrt{6}}{6}
=\frac{6\sqrt{6}}{6}
=\sqrt{6}
$