1260=(16+2x)(28+2x)

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Solution for 1260=(16+2x)(28+2x) equation:



1260=(16+2x)(28+2x)
We move all terms to the left:
1260-((16+2x)(28+2x))=0
We add all the numbers together, and all the variables
-((2x+16)(2x+28))+1260=0
We multiply parentheses ..
-((+4x^2+56x+32x+448))+1260=0
We calculate terms in parentheses: -((+4x^2+56x+32x+448)), so:
(+4x^2+56x+32x+448)
We get rid of parentheses
4x^2+56x+32x+448
We add all the numbers together, and all the variables
4x^2+88x+448
Back to the equation:
-(4x^2+88x+448)
We get rid of parentheses
-4x^2-88x-448+1260=0
We add all the numbers together, and all the variables
-4x^2-88x+812=0
a = -4; b = -88; c = +812;
Δ = b2-4ac
Δ = -882-4·(-4)·812
Δ = 20736
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{20736}=144$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-88)-144}{2*-4}=\frac{-56}{-8} =+7 $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-88)+144}{2*-4}=\frac{232}{-8} =-29 $

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