16(4-3x)=96(-x2+1)

Solution for 16(4-3x)=96(-x2+1) equation:

16(4-3x)=96(-x2+1)

We move all terms to the left:

16(4-3x)-(96(-x2+1))=0

We add all the numbers together, and all the variables

-(96(-1x^2+1))+16(-3x+4)=0

We multiply parentheses

-(96(-1x^2+1))-48x+64=0


We calculate terms in parentheses: -(96(-1x^2+1)), so:

96(-1x^2+1)

We multiply parentheses

-96x^2+96

Back to the equation:

-(-96x^2+96)


We get rid of parentheses

96x^2-48x-96+64=0

We add all the numbers together, and all the variables

96x^2-48x-32=0

a = 96; b = -48; c = -32;
Δ = b2-4ac
Δ = -482-4·96·(-32)
Δ = 14592
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:

$\sqrt{\Delta}=\sqrt{14592}=\sqrt{256*57}=\sqrt{256}*\sqrt{57}=16\sqrt{57}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-48)-16\sqrt{57}}{2*96}=\frac{48-16\sqrt{57}}{192}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-48)+16\sqrt{57}}{2*96}=\frac{48+16\sqrt{57}}{192}
$