(2x^2-68x+11)/2x+3=1

Solution for (2x^2-68x+11)/2x+3=1 equation:

(2x^2-68x+11)/2x+3=1

We move all terms to the left:

(2x^2-68x+11)/2x+3-(1)=0


Domain of the equation: 2x!=0

x!=0/2

x!=0

x∈R


We add all the numbers together, and all the variables

(2x^2-68x+11)/2x+2=0

We multiply all the terms by the denominator


(2x^2-68x+11)+2*2x=0

Wy multiply elements

(2x^2-68x+11)+4x=0

We get rid of parentheses

2x^2-68x+4x+11=0

We add all the numbers together, and all the variables

2x^2-64x+11=0

a = 2; b = -64; c = +11;
Δ = b2-4ac
Δ = -642-4·2·11
Δ = 4008
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:

$\sqrt{\Delta}=\sqrt{4008}=\sqrt{4*1002}=\sqrt{4}*\sqrt{1002}=2\sqrt{1002}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-64)-2\sqrt{1002}}{2*2}=\frac{64-2\sqrt{1002}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-64)+2\sqrt{1002}}{2*2}=\frac{64+2\sqrt{1002}}{4}
$