6x+x(x-5)=3(2-3x)

Solution for 6x+x(x-5)=3(2-3x) equation:

6x+x(x-5)=3(2-3x)

We move all terms to the left:

6x+x(x-5)-(3(2-3x))=0

We add all the numbers together, and all the variables

6x+x(x-5)-(3(-3x+2))=0

We multiply parentheses

x^2+6x-5x-(3(-3x+2))=0


We calculate terms in parentheses: -(3(-3x+2)), so:

3(-3x+2)

We multiply parentheses

-9x+6

Back to the equation:

-(-9x+6)


We add all the numbers together, and all the variables

x^2+x-(-9x+6)=0

We get rid of parentheses

x^2+x+9x-6=0

We add all the numbers together, and all the variables

x^2+10x-6=0

a = 1; b = 10; c = -6;
Δ = b2-4ac
Δ = 102-4·1·(-6)
Δ = 124
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{124}=\sqrt{4*31}=\sqrt{4}*\sqrt{31}=2\sqrt{31}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{31}}{2*1}=\frac{-10-2\sqrt{31}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{31}}{2*1}=\frac{-10+2\sqrt{31}}{2}
$