(1/4)+(1/5)x=10

Solution for (1/4)+(1/5)x=10 equation:

(1/4)+(1/5)x=10

We move all terms to the left:

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


Domain of the equation: 5)x!=0

x!=0/1

x!=0

x∈R


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

We add all the numbers together, and all the variables

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

We multiply parentheses

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

We get rid of parentheses

x^2-10+1/4=0

We multiply all the terms by the denominator


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

We add all the numbers together, and all the variables

x^2*4-39=0

Wy multiply elements

4x^2-39=0

a = 4; b = 0; c = -39;
Δ = b2-4ac
Δ = 02-4·4·(-39)
Δ = 624
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:

$\sqrt{\Delta}=\sqrt{624}=\sqrt{16*39}=\sqrt{16}*\sqrt{39}=4\sqrt{39}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{39}}{2*4}=\frac{0-4\sqrt{39}}{8}
=-\frac{4\sqrt{39}}{8}
=-\frac{\sqrt{39}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{39}}{2*4}=\frac{0+4\sqrt{39}}{8}
=\frac{4\sqrt{39}}{8}
=\frac{\sqrt{39}}{2}
$