126=(x+3)(2x+10)

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Solution for 126=(x+3)(2x+10) equation:



126=(x+3)(2x+10)
We move all terms to the left:
126-((x+3)(2x+10))=0
We multiply parentheses ..
-((+2x^2+10x+6x+30))+126=0
We calculate terms in parentheses: -((+2x^2+10x+6x+30)), so:
(+2x^2+10x+6x+30)
We get rid of parentheses
2x^2+10x+6x+30
We add all the numbers together, and all the variables
2x^2+16x+30
Back to the equation:
-(2x^2+16x+30)
We get rid of parentheses
-2x^2-16x-30+126=0
We add all the numbers together, and all the variables
-2x^2-16x+96=0
a = -2; b = -16; c = +96;
Δ = b2-4ac
Δ = -162-4·(-2)·96
Δ = 1024
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1024}=32$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-32}{2*-2}=\frac{-16}{-4} =+4 $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+32}{2*-2}=\frac{48}{-4} =-12 $

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