(3x+2)(x-4)=x(x+5)-21

Solution for (3x+2)(x-4)=x(x+5)-21 equation:

(3x+2)(x-4)=x(x+5)-21

We move all terms to the left:

(3x+2)(x-4)-(x(x+5)-21)=0

We multiply parentheses ..

(+3x^2-12x+2x-8)-(x(x+5)-21)=0


We calculate terms in parentheses: -(x(x+5)-21), so:

x(x+5)-21

We multiply parentheses

x^2+5x-21

Back to the equation:

-(x^2+5x-21)


We get rid of parentheses

3x^2-x^2-12x+2x-5x-8+21=0

We add all the numbers together, and all the variables

2x^2-15x+13=0

a = 2; b = -15; c = +13;
Δ = b2-4ac
Δ = -152-4·2·13
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-15)-11}{2*2}=\frac{4}{4}
=1
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+11}{2*2}=\frac{26}{4}
=6+1/2
$