2x^2-x-6=(2x-3)(2x-3)

Solution for 2x^2-x-6=(2x-3)(2x-3) equation:

2x^2-x-6=(2x-3)(2x-3)

We move all terms to the left:

2x^2-x-6-((2x-3)(2x-3))=0

We add all the numbers together, and all the variables

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

We multiply parentheses ..

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


We calculate terms in parentheses: -((+4x^2-6x-6x+9)), so:

(+4x^2-6x-6x+9)

We get rid of parentheses

4x^2-6x-6x+9

We add all the numbers together, and all the variables

4x^2-12x+9

Back to the equation:

-(4x^2-12x+9)


We add all the numbers together, and all the variables

2x^2-1x-(4x^2-12x+9)-6=0

We get rid of parentheses

2x^2-4x^2-1x+12x-9-6=0

We add all the numbers together, and all the variables

-2x^2+11x-15=0

a = -2; b = 11; c = -15;
Δ = b2-4ac
Δ = 112-4·(-2)·(-15)
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-1}{2*-2}=\frac{-12}{-4}
=+3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+1}{2*-2}=\frac{-10}{-4}
=2+1/2
$