8=18+(3/8)(16-40x)

Solution for 8=18+(3/8)(16-40x) equation:

8=18+(3/8)(16-40x)

We move all terms to the left:

8-(18+(3/8)(16-40x))=0


Domain of the equation: 8)(16-40x))!=0

x∈R


We add all the numbers together, and all the variables

-(18+(+3/8)(-40x+16))+8=0

We multiply parentheses ..

-(18+(-120x^2+3/8*16))+8=0

We multiply all the terms by the denominator


-(18+(-120x^2+3+8*8*16))=0


We calculate terms in parentheses: -(18+(-120x^2+3+8*8*16)), so:

18+(-120x^2+3+8*8*16)

determiningTheFunctionDomain
(-120x^2+3+8*8*16)+18

We get rid of parentheses

-120x^2+3+18+8*8*16

We add all the numbers together, and all the variables

-120x^2+1045

Back to the equation:

-(-120x^2+1045)


We get rid of parentheses

120x^2-1045=0

a = 120; b = 0; c = -1045;
Δ = b2-4ac
Δ = 02-4·120·(-1045)
Δ = 501600
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:

$\sqrt{\Delta}=\sqrt{501600}=\sqrt{400*1254}=\sqrt{400}*\sqrt{1254}=20\sqrt{1254}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-20\sqrt{1254}}{2*120}=\frac{0-20\sqrt{1254}}{240}
=-\frac{20\sqrt{1254}}{240}
=-\frac{\sqrt{1254}}{12}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+20\sqrt{1254}}{2*120}=\frac{0+20\sqrt{1254}}{240}
=\frac{20\sqrt{1254}}{240}
=\frac{\sqrt{1254}}{12}
$