5+8(3x^2+2)=23-2(2x-5)

Solution for 5+8(3x^2+2)=23-2(2x-5) equation:

5+8(3x^2+2)=23-2(2x-5)

We move all terms to the left:

5+8(3x^2+2)-(23-2(2x-5))=0

We multiply parentheses

24x^2-(23-2(2x-5))+16+5=0


We calculate terms in parentheses: -(23-2(2x-5)), so:

23-2(2x-5)

determiningTheFunctionDomain
-2(2x-5)+23

We multiply parentheses

-4x+10+23

We add all the numbers together, and all the variables

-4x+33

Back to the equation:

-(-4x+33)


We add all the numbers together, and all the variables

24x^2-(-4x+33)+21=0

We get rid of parentheses

24x^2+4x-33+21=0

We add all the numbers together, and all the variables

24x^2+4x-12=0

a = 24; b = 4; c = -12;
Δ = b2-4ac
Δ = 42-4·24·(-12)
Δ = 1168
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:

$\sqrt{\Delta}=\sqrt{1168}=\sqrt{16*73}=\sqrt{16}*\sqrt{73}=4\sqrt{73}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{73}}{2*24}=\frac{-4-4\sqrt{73}}{48}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{73}}{2*24}=\frac{-4+4\sqrt{73}}{48}
$