8x-(4x^2)+(40/3)=0

Solution for 8x-(4x^2)+(40/3)=0 equation:

8x-(4x^2)+(40/3)=0

determiningTheFunctionDomain
-4x^2+8x+(40/3)=0

We add all the numbers together, and all the variables

-4x^2+8x+(+40/3)=0

We get rid of parentheses

-4x^2+8x+40/3=0

We multiply all the terms by the denominator


-4x^2*3+8x*3+40=0

Wy multiply elements

-12x^2+24x+40=0

a = -12; b = 24; c = +40;
Δ = b2-4ac
Δ = 242-4·(-12)·40
Δ = 2496
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:

$\sqrt{\Delta}=\sqrt{2496}=\sqrt{64*39}=\sqrt{64}*\sqrt{39}=8\sqrt{39}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-8\sqrt{39}}{2*-12}=\frac{-24-8\sqrt{39}}{-24}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+8\sqrt{39}}{2*-12}=\frac{-24+8\sqrt{39}}{-24}
$