(8/3)(1x+6)=14

Solution for (8/3)(1x+6)=14 equation:

(8/3)(1x+6)=14

We move all terms to the left:

(8/3)(1x+6)-(14)=0


Domain of the equation: 3)(1x+6)!=0

x∈R


We add all the numbers together, and all the variables

(+8/3)(x+6)-14=0

We multiply parentheses ..

(+8x^2+8/3*6)-14=0

We multiply all the terms by the denominator


(+8x^2+8-14*3*6)=0

We get rid of parentheses

8x^2+8-14*3*6=0

We add all the numbers together, and all the variables

8x^2-244=0

a = 8; b = 0; c = -244;
Δ = b2-4ac
Δ = 02-4·8·(-244)
Δ = 7808
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:

$\sqrt{\Delta}=\sqrt{7808}=\sqrt{64*122}=\sqrt{64}*\sqrt{122}=8\sqrt{122}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{122}}{2*8}=\frac{0-8\sqrt{122}}{16}
=-\frac{8\sqrt{122}}{16}
=-\frac{\sqrt{122}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{122}}{2*8}=\frac{0+8\sqrt{122}}{16}
=\frac{8\sqrt{122}}{16}
=\frac{\sqrt{122}}{2}
$