5-2x+6x=(1/2)(10+8x)

Solution for 5-2x+6x=(1/2)(10+8x) equation:

5-2x+6x=(1/2)(10+8x)

We move all terms to the left:

5-2x+6x-((1/2)(10+8x))=0


Domain of the equation: 2)(10+8x))!=0

x∈R


We add all the numbers together, and all the variables

-2x+6x-((+1/2)(8x+10))+5=0

We add all the numbers together, and all the variables

4x-((+1/2)(8x+10))+5=0

We multiply parentheses ..

-((+8x^2+1/2*10))+4x+5=0

We multiply all the terms by the denominator


-((+8x^2+1+4x*2*10))+5*2*10))=0


We calculate terms in parentheses: -((+8x^2+1+4x*2*10)), so:

(+8x^2+1+4x*2*10)

We get rid of parentheses

8x^2+4x*2*10+1

Wy multiply elements

8x^2+80x*1+1

Wy multiply elements

8x^2+80x+1

Back to the equation:

-(8x^2+80x+1)


We add all the numbers together, and all the variables

-(8x^2+80x+1)=0

We get rid of parentheses

-8x^2-80x-1=0

a = -8; b = -80; c = -1;
Δ = b2-4ac
Δ = -802-4·(-8)·(-1)
Δ = 6368
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:

$\sqrt{\Delta}=\sqrt{6368}=\sqrt{16*398}=\sqrt{16}*\sqrt{398}=4\sqrt{398}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-80)-4\sqrt{398}}{2*-8}=\frac{80-4\sqrt{398}}{-16}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-80)+4\sqrt{398}}{2*-8}=\frac{80+4\sqrt{398}}{-16}
$