(4x^2-5x)=(7x-10x^2)

Solution for (4x^2-5x)=(7x-10x^2) equation:

(4x^2-5x)=(7x-10x^2)

We move all terms to the left:

(4x^2-5x)-((7x-10x^2))=0

We get rid of parentheses

-((7x-10x^2))+4x^2-5x=0


We calculate terms in parentheses: -((7x-10x^2)), so:

(7x-10x^2)

We get rid of parentheses

-10x^2+7x

Back to the equation:

-(-10x^2+7x)


We add all the numbers together, and all the variables

4x^2-(-10x^2+7x)-5x=0

We get rid of parentheses

4x^2+10x^2-7x-5x=0

We add all the numbers together, and all the variables

14x^2-12x=0

a = 14; b = -12; c = 0;
Δ = b2-4ac
Δ = -122-4·14·0
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-12}{2*14}=\frac{0}{28}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+12}{2*14}=\frac{24}{28}
=6/7
$