(2x/7)=2x^2

Solution for (2x/7)=2x^2 equation:

(2x/7)=2x^2

We move all terms to the left:

(2x/7)-(2x^2)=0

determiningTheFunctionDomain
-2x^2+(2x/7)=0

We add all the numbers together, and all the variables

-2x^2+(+2x/7)=0

We get rid of parentheses

-2x^2+2x/7=0

We multiply all the terms by the denominator


-2x^2*7+2x=0

Wy multiply elements

-14x^2+2x=0

a = -14; b = 2; c = 0;
Δ = b2-4ac
Δ = 22-4·(-14)·0
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2}{2*-14}=\frac{-4}{-28}
=1/7
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2}{2*-14}=\frac{0}{-28}
=0
$