(71/5)(2x)=470

Solution for (71/5)(2x)=470 equation:

(71/5)(2x)=470

We move all terms to the left:

(71/5)(2x)-(470)=0


Domain of the equation: 5)2x!=0

x!=0/1

x!=0

x∈R


We add all the numbers together, and all the variables

(+71/5)2x-470=0

We multiply parentheses

142x^2-470=0

a = 142; b = 0; c = -470;
Δ = b2-4ac
Δ = 02-4·142·(-470)
Δ = 266960
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:

$\sqrt{\Delta}=\sqrt{266960}=\sqrt{16*16685}=\sqrt{16}*\sqrt{16685}=4\sqrt{16685}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{16685}}{2*142}=\frac{0-4\sqrt{16685}}{284}
=-\frac{4\sqrt{16685}}{284}
=-\frac{\sqrt{16685}}{71}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{16685}}{2*142}=\frac{0+4\sqrt{16685}}{284}
=\frac{4\sqrt{16685}}{284}
=\frac{\sqrt{16685}}{71}
$