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

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

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

We move all terms to the left:

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


Domain of the equation: 5)x!=0

x!=0/1

x!=0

x∈R


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

We add all the numbers together, and all the variables

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

We multiply parentheses

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

We get rid of parentheses

4x^2-10+4/5=0

We multiply all the terms by the denominator


4x^2*5+4-10*5=0

We add all the numbers together, and all the variables

4x^2*5-46=0

Wy multiply elements

20x^2-46=0

a = 20; b = 0; c = -46;
Δ = b2-4ac
Δ = 02-4·20·(-46)
Δ = 3680
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:

$\sqrt{\Delta}=\sqrt{3680}=\sqrt{16*230}=\sqrt{16}*\sqrt{230}=4\sqrt{230}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{230}}{2*20}=\frac{0-4\sqrt{230}}{40}
=-\frac{4\sqrt{230}}{40}
=-\frac{\sqrt{230}}{10}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{230}}{2*20}=\frac{0+4\sqrt{230}}{40}
=\frac{4\sqrt{230}}{40}
=\frac{\sqrt{230}}{10}
$