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

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

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

We move all terms to the left:

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

We multiply parentheses ..

(+4x^2-20x+3x-15)-((2x-3)(x-2)+x)=0


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

(2x-3)(x-2)+x

We add all the numbers together, and all the variables

x+(2x-3)(x-2)

We multiply parentheses ..

(+2x^2-4x-3x+6)+x

We get rid of parentheses

2x^2-4x-3x+x+6

We add all the numbers together, and all the variables

2x^2-6x+6

Back to the equation:

-(2x^2-6x+6)


We get rid of parentheses

4x^2-2x^2-20x+3x+6x-15-6=0

We add all the numbers together, and all the variables

2x^2-11x-21=0

a = 2; b = -11; c = -21;
Δ = b2-4ac
Δ = -112-4·2·(-21)
Δ = 289
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{289}=17$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11)-17}{2*2}=\frac{-6}{4}
=-1+1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+17}{2*2}=\frac{28}{4}
=7
$