(3/4)q-7=14

Solution for (3/4)q-7=14 equation:

(3/4)q-7=14

We move all terms to the left:

(3/4)q-7-(14)=0


Domain of the equation: 4)q!=0

q!=0/1

q!=0

q∈R


We add all the numbers together, and all the variables

(+3/4)q-7-14=0

We add all the numbers together, and all the variables

(+3/4)q-21=0

We multiply parentheses

3q^2-21=0

a = 3; b = 0; c = -21;
Δ = b2-4ac
Δ = 02-4·3·(-21)
Δ = 252
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{252}=\sqrt{36*7}=\sqrt{36}*\sqrt{7}=6\sqrt{7}$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6\sqrt{7}}{2*3}=\frac{0-6\sqrt{7}}{6}
=-\frac{6\sqrt{7}}{6}
=-\sqrt{7}
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6\sqrt{7}}{2*3}=\frac{0+6\sqrt{7}}{6}
=\frac{6\sqrt{7}}{6}
=\sqrt{7}
$