1/4q+8=2q-16

Solution for 1/4q+8=2q-16 equation:

1/4q+8=2q-16

We move all terms to the left:

1/4q+8-(2q-16)=0


Domain of the equation: 4q!=0

q!=0/4

q!=0

q∈R


We get rid of parentheses

1/4q-2q+16+8=0

We multiply all the terms by the denominator


-2q*4q+16*4q+8*4q+1=0

Wy multiply elements

-8q^2+64q+32q+1=0

We add all the numbers together, and all the variables

-8q^2+96q+1=0

a = -8; b = 96; c = +1;
Δ = b2-4ac
Δ = 962-4·(-8)·1
Δ = 9248
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:

$\sqrt{\Delta}=\sqrt{9248}=\sqrt{4624*2}=\sqrt{4624}*\sqrt{2}=68\sqrt{2}$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(96)-68\sqrt{2}}{2*-8}=\frac{-96-68\sqrt{2}}{-16}
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(96)+68\sqrt{2}}{2*-8}=\frac{-96+68\sqrt{2}}{-16}
$