-18+(1/7)(42x+7)-3x=16

Solution for -18+(1/7)(42x+7)-3x=16 equation:

-18+(1/7)(42x+7)-3x=16

We move all terms to the left:

-18+(1/7)(42x+7)-3x-(16)=0


Domain of the equation: 7)(42x+7)!=0

x∈R


We add all the numbers together, and all the variables

(+1/7)(42x+7)-3x-18-16=0

We add all the numbers together, and all the variables

-3x+(+1/7)(42x+7)-34=0

We multiply parentheses ..

(+42x^2+1/7*7)-3x-34=0

We multiply all the terms by the denominator


(+42x^2+1-3x*7*7)-34*7*7)=0

We add all the numbers together, and all the variables

(+42x^2+1-3x*7*7)=0

We get rid of parentheses

42x^2-3x*7*7+1=0

Wy multiply elements

42x^2-147x*7+1=0

Wy multiply elements

42x^2-1029x+1=0

a = 42; b = -1029; c = +1;
Δ = b2-4ac
Δ = -10292-4·42·1
Δ = 1058673
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1029)-\sqrt{1058673}}{2*42}=\frac{1029-\sqrt{1058673}}{84}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1029)+\sqrt{1058673}}{2*42}=\frac{1029+\sqrt{1058673}}{84}
$