(1/4)(7x+2)=6

Solution for (1/4)(7x+2)=6 equation:

(1/4)(7x+2)=6

We move all terms to the left:

(1/4)(7x+2)-(6)=0


Domain of the equation: 4)(7x+2)!=0

x∈R


We add all the numbers together, and all the variables

(+1/4)(7x+2)-6=0

We multiply parentheses ..

(+7x^2+1/4*2)-6=0

We multiply all the terms by the denominator


(+7x^2+1-6*4*2)=0

We get rid of parentheses

7x^2+1-6*4*2=0

We add all the numbers together, and all the variables

7x^2-47=0

a = 7; b = 0; c = -47;
Δ = b2-4ac
Δ = 02-4·7·(-47)
Δ = 1316
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:

$\sqrt{\Delta}=\sqrt{1316}=\sqrt{4*329}=\sqrt{4}*\sqrt{329}=2\sqrt{329}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{329}}{2*7}=\frac{0-2\sqrt{329}}{14}
=-\frac{2\sqrt{329}}{14}
=-\frac{\sqrt{329}}{7}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{329}}{2*7}=\frac{0+2\sqrt{329}}{14}
=\frac{2\sqrt{329}}{14}
=\frac{\sqrt{329}}{7}
$