1/2x^+4x=10

Solution for 1/2x^+4x=10 equation:

1/2x^+4x=10

We move all terms to the left:

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


Domain of the equation: 2x^!=0

x!=0/2

x!=0

x∈R


We add all the numbers together, and all the variables

4x+1/2x^-10=0

We multiply all the terms by the denominator


4x*2x^-10*2x^+1=0

Wy multiply elements

8x^2-20x+1=0

a = 8; b = -20; c = +1;
Δ = b2-4ac
Δ = -202-4·8·1
Δ = 368
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:

$\sqrt{\Delta}=\sqrt{368}=\sqrt{16*23}=\sqrt{16}*\sqrt{23}=4\sqrt{23}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-4\sqrt{23}}{2*8}=\frac{20-4\sqrt{23}}{16}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+4\sqrt{23}}{2*8}=\frac{20+4\sqrt{23}}{16}
$