(4x+5)+(3x+6)=4x(5+3x)+6

Solution for (4x+5)+(3x+6)=4x(5+3x)+6 equation:

(4x+5)+(3x+6)=4x(5+3x)+6

We move all terms to the left:

(4x+5)+(3x+6)-(4x(5+3x)+6)=0

We add all the numbers together, and all the variables

(4x+5)+(3x+6)-(4x(3x+5)+6)=0

We get rid of parentheses

4x+3x-(4x(3x+5)+6)+5+6=0


We calculate terms in parentheses: -(4x(3x+5)+6), so:

4x(3x+5)+6

We multiply parentheses

12x^2+20x+6

Back to the equation:

-(12x^2+20x+6)


We add all the numbers together, and all the variables

7x-(12x^2+20x+6)+11=0

We get rid of parentheses

-12x^2+7x-20x-6+11=0

We add all the numbers together, and all the variables

-12x^2-13x+5=0

a = -12; b = -13; c = +5;
Δ = b2-4ac
Δ = -132-4·(-12)·5
Δ = 409
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-13)-\sqrt{409}}{2*-12}=\frac{13-\sqrt{409}}{-24}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+\sqrt{409}}{2*-12}=\frac{13+\sqrt{409}}{-24}
$