(1/4x)+7=(2x)-3

Solution for (1/4x)+7=(2x)-3 equation:

(1/4x)+7=(2x)-3

We move all terms to the left:

(1/4x)+7-((2x)-3)=0


Domain of the equation: 4x)!=0

x!=0/1

x!=0

x∈R


We add all the numbers together, and all the variables

(+1/4x)-(2x-3)+7=0

We get rid of parentheses

1/4x-2x+3+7=0

We multiply all the terms by the denominator


-2x*4x+3*4x+7*4x+1=0

Wy multiply elements

-8x^2+12x+28x+1=0

We add all the numbers together, and all the variables

-8x^2+40x+1=0

a = -8; b = 40; c = +1;
Δ = b2-4ac
Δ = 402-4·(-8)·1
Δ = 1632
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1632}=\sqrt{16*102}=\sqrt{16}*\sqrt{102}=4\sqrt{102}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-4\sqrt{102}}{2*-8}=\frac{-40-4\sqrt{102}}{-16}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+4\sqrt{102}}{2*-8}=\frac{-40+4\sqrt{102}}{-16}
$