-3x+4x=(1x+-3x)(1x+-3x)

Solution for -3x+4x=(1x+-3x)(1x+-3x) equation:

-3x+4x=(1x+-3x)(1x+-3x)

We move all terms to the left:

-3x+4x-((1x+-3x)(1x+-3x))=0

We add all the numbers together, and all the variables

-3x+4x-((-2x)(-2x))=0

We add all the numbers together, and all the variables

x-((-2x)(-2x))=0

We multiply parentheses ..

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


We calculate terms in parentheses: -((+4x^2)), so:

(+4x^2)

We get rid of parentheses

4x^2

Back to the equation:

-(4x^2)


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

$\sqrt{\Delta}=\sqrt{1}=1$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-1}{2*-4}=\frac{-2}{-8}
=1/4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+1}{2*-4}=\frac{0}{-8}
=0
$