126=(4x+1)(x+4)

Solution for 126=(4x+1)(x+4) equation:

126=(4x+1)(x+4)

We move all terms to the left:

126-((4x+1)(x+4))=0

We multiply parentheses ..

-((+4x^2+16x+x+4))+126=0


We calculate terms in parentheses: -((+4x^2+16x+x+4)), so:

(+4x^2+16x+x+4)

We get rid of parentheses

4x^2+16x+x+4

We add all the numbers together, and all the variables

4x^2+17x+4

Back to the equation:

-(4x^2+17x+4)


We get rid of parentheses

-4x^2-17x-4+126=0

We add all the numbers together, and all the variables

-4x^2-17x+122=0

a = -4; b = -17; c = +122;
Δ = b2-4ac
Δ = -172-4·(-4)·122
Δ = 2241
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{2241}=\sqrt{9*249}=\sqrt{9}*\sqrt{249}=3\sqrt{249}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-17)-3\sqrt{249}}{2*-4}=\frac{17-3\sqrt{249}}{-8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-17)+3\sqrt{249}}{2*-4}=\frac{17+3\sqrt{249}}{-8}
$