(3x+10)(x-3)=x(2x+1)+6

Solution for (3x+10)(x-3)=x(2x+1)+6 equation:

(3x+10)(x-3)=x(2x+1)+6

We move all terms to the left:

(3x+10)(x-3)-(x(2x+1)+6)=0

We multiply parentheses ..

(+3x^2-9x+10x-30)-(x(2x+1)+6)=0


We calculate terms in parentheses: -(x(2x+1)+6), so:

x(2x+1)+6

We multiply parentheses

2x^2+x+6

Back to the equation:

-(2x^2+x+6)


We get rid of parentheses

3x^2-2x^2-9x+10x-x-30-6=0

We add all the numbers together, and all the variables

x^2-36=0

a = 1; b = 0; c = -36;
Δ = b2-4ac
Δ = 02-4·1·(-36)
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{-(0)-12}{2*1}=\frac{-12}{2}
=-6
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+12}{2*1}=\frac{12}{2}
=6
$