3x+3=42/2x+10

Solution for 3x+3=42/2x+10 equation:

3x+3=42/2x+10

We move all terms to the left:

3x+3-(42/2x+10)=0


Domain of the equation: 2x+10)!=0

x∈R


We get rid of parentheses

3x-42/2x-10+3=0

We multiply all the terms by the denominator


3x*2x-10*2x+3*2x-42=0

Wy multiply elements

6x^2-20x+6x-42=0

We add all the numbers together, and all the variables

6x^2-14x-42=0

a = 6; b = -14; c = -42;
Δ = b2-4ac
Δ = -142-4·6·(-42)
Δ = 1204
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:

$\sqrt{\Delta}=\sqrt{1204}=\sqrt{4*301}=\sqrt{4}*\sqrt{301}=2\sqrt{301}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-2\sqrt{301}}{2*6}=\frac{14-2\sqrt{301}}{12}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+2\sqrt{301}}{2*6}=\frac{14+2\sqrt{301}}{12}
$