2x^-11x-21=(2x-3)(x+7)

Solution for 2x^-11x-21=(2x-3)(x+7) equation:

2x^-11x-21=(2x-3)(x+7)

We move all terms to the left:

2x^-11x-21-((2x-3)(x+7))=0

We add all the numbers together, and all the variables

-9x-((2x-3)(x+7))-21=0

We multiply parentheses ..

-((+2x^2+14x-3x-21))-9x-21=0


We calculate terms in parentheses: -((+2x^2+14x-3x-21)), so:

(+2x^2+14x-3x-21)

We get rid of parentheses

2x^2+14x-3x-21

We add all the numbers together, and all the variables

2x^2+11x-21

Back to the equation:

-(2x^2+11x-21)


We add all the numbers together, and all the variables

-9x-(2x^2+11x-21)-21=0

We get rid of parentheses

-2x^2-9x-11x+21-21=0

We add all the numbers together, and all the variables

-2x^2-20x=0

a = -2; b = -20; c = 0;
Δ = b2-4ac
Δ = -202-4·(-2)·0
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-20}{2*-2}=\frac{0}{-4}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+20}{2*-2}=\frac{40}{-4}
=-10
$