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

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
$