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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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


We calculate terms in parentheses: +(3x-1x(x-4)), so:

3x-1x(x-4)

We multiply parentheses

-x^2+3x+4x

We add all the numbers together, and all the variables

-1x^2+7x

Back to the equation:

+(-1x^2+7x)


We get rid of parentheses

-1x^2+7x=0

a = -1; b = 7; c = 0;
Δ = b2-4ac
Δ = 72-4·(-1)·0
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-7}{2*-1}=\frac{-14}{-2}
=+7
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+7}{2*-1}=\frac{0}{-2}
=0
$