(3x+1)(2x-1)-2x2=(2x-3)2+6x+5

Solution for (3x+1)(2x-1)-2x2=(2x-3)2+6x+5 equation:

(3x+1)(2x-1)-2x^2=(2x-3)2+6x+5

We move all terms to the left:

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

We multiply parentheses ..

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


We calculate terms in parentheses: -((2x-3)2+6x+5), so:

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

We add all the numbers together, and all the variables

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

We multiply parentheses

6x+4x-6+5

We add all the numbers together, and all the variables

10x-1

Back to the equation:

-(10x-1)


We get rid of parentheses

-2x^2+6x^2-3x+2x-10x-1+1=0

We add all the numbers together, and all the variables

4x^2-11x=0

a = 4; b = -11; c = 0;
Δ = b2-4ac
Δ = -112-4·4·0
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{-(-11)-11}{2*4}=\frac{0}{8}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+11}{2*4}=\frac{22}{8}
=2+3/4
$