(7x2+4x)=(2x-4x)

Solution for (7x2+4x)=(2x-4x) equation:

(7x^2+4x)=(2x-4x)

We move all terms to the left:

(7x^2+4x)-((2x-4x))=0

We add all the numbers together, and all the variables

(7x^2+4x)-((-2x))=0

We get rid of parentheses

7x^2+4x-((-2x))=0


We calculate terms in parentheses: -((-2x)), so:

(-2x)

We get rid of parentheses

-2x

Back to the equation:

-(-2x)


We get rid of parentheses

7x^2+4x+2x=0

We add all the numbers together, and all the variables

7x^2+6x=0

a = 7; b = 6; c = 0;
Δ = b2-4ac
Δ = 62-4·7·0
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-6}{2*7}=\frac{-12}{14}
=-6/7
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+6}{2*7}=\frac{0}{14}
=0
$