(24x^2+48x+20x+40)/2=A

Solution for (24x^2+48x+20x+40)/2=A equation:

(24x^2+48x+20x+40)/2=

We move all terms to the left:

(24x^2+48x+20x+40)/2-()=0

We add all the numbers together, and all the variables

(24x^2+48x+20x+40)/2=0

We multiply all the terms by the denominator


(24x^2+48x+20x+40)=0

We get rid of parentheses

24x^2+48x+20x+40=0

We add all the numbers together, and all the variables

24x^2+68x+40=0

a = 24; b = 68; c = +40;
Δ = b2-4ac
Δ = 682-4·24·40
Δ = 784
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{784}=28$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(68)-28}{2*24}=\frac{-96}{48}
=-2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(68)+28}{2*24}=\frac{-40}{48}
=-5/6
$