16=(x+2)(2x-3)

Solution for 16=(x+2)(2x-3) equation:

16=(x+2)(2x-3)

We move all terms to the left:

16-((x+2)(2x-3))=0

We multiply parentheses ..

-((+2x^2-3x+4x-6))+16=0


We calculate terms in parentheses: -((+2x^2-3x+4x-6)), so:

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

2x^2+x-6

Back to the equation:

-(2x^2+x-6)


We get rid of parentheses

-2x^2-x+6+16=0

We add all the numbers together, and all the variables

-2x^2-1x+22=0

a = -2; b = -1; c = +22;
Δ = b2-4ac
Δ = -12-4·(-2)·22
Δ = 177
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-\sqrt{177}}{2*-2}=\frac{1-\sqrt{177}}{-4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{177}}{2*-2}=\frac{1+\sqrt{177}}{-4}
$