A=(3x+1)(x-4)+11x

Solution for A=(3x+1)(x-4)+11x equation:

=(3A+1)(A-4)+11A

We move all terms to the left:

-((3A+1)(A-4)+11A)=0

We multiply parentheses ..

-((+3A^2-12A+A-4)+11A)=0


We calculate terms in parentheses: -((+3A^2-12A+A-4)+11A), so:

(+3A^2-12A+A-4)+11A

We get rid of parentheses

3A^2-12A+A+11A-4

We add all the numbers together, and all the variables

3A^2-4

Back to the equation:

-(3A^2-4)


We get rid of parentheses

-3A^2+4=0

a = -3; b = 0; c = +4;
Δ = b2-4ac
Δ = 02-4·(-3)·4
Δ = 48
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$A_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$A_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{48}=\sqrt{16*3}=\sqrt{16}*\sqrt{3}=4\sqrt{3}$
$A_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{3}}{2*-3}=\frac{0-4\sqrt{3}}{-6}
=-\frac{4\sqrt{3}}{-6}
=-\frac{2\sqrt{3}}{-3}
$
$A_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{3}}{2*-3}=\frac{0+4\sqrt{3}}{-6}
=\frac{4\sqrt{3}}{-6}
=\frac{2\sqrt{3}}{-3}
$