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

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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} $

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