40=(4x+3)(6x-3)

Solution for 40=(4x+3)(6x-3) equation:

40=(4x+3)(6x-3)

We move all terms to the left:

40-((4x+3)(6x-3))=0

We multiply parentheses ..

-((+24x^2-12x+18x-9))+40=0


We calculate terms in parentheses: -((+24x^2-12x+18x-9)), so:

(+24x^2-12x+18x-9)

We get rid of parentheses

24x^2-12x+18x-9

We add all the numbers together, and all the variables

24x^2+6x-9

Back to the equation:

-(24x^2+6x-9)


We get rid of parentheses

-24x^2-6x+9+40=0

We add all the numbers together, and all the variables

-24x^2-6x+49=0

a = -24; b = -6; c = +49;
Δ = b2-4ac
Δ = -62-4·(-24)·49
Δ = 4740
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:

$\sqrt{\Delta}=\sqrt{4740}=\sqrt{4*1185}=\sqrt{4}*\sqrt{1185}=2\sqrt{1185}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-2\sqrt{1185}}{2*-24}=\frac{6-2\sqrt{1185}}{-48}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+2\sqrt{1185}}{2*-24}=\frac{6+2\sqrt{1185}}{-48}
$