(3x+4)(2x+1)=5(x-1)+9

Solution for (3x+4)(2x+1)=5(x-1)+9 equation:

(3x+4)(2x+1)=5(x-1)+9

We move all terms to the left:

(3x+4)(2x+1)-(5(x-1)+9)=0

We multiply parentheses ..

(+6x^2+3x+8x+4)-(5(x-1)+9)=0


We calculate terms in parentheses: -(5(x-1)+9), so:

5(x-1)+9

We multiply parentheses

5x-5+9

We add all the numbers together, and all the variables

5x+4

Back to the equation:

-(5x+4)


We get rid of parentheses

6x^2+3x+8x-5x+4-4=0

We add all the numbers together, and all the variables

6x^2+6x=0

a = 6; b = 6; c = 0;
Δ = b2-4ac
Δ = 62-4·6·0
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-6}{2*6}=\frac{-12}{12}
=-1
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+6}{2*6}=\frac{0}{12}
=0
$