a/5a-4=3a+10

Solution for a/5a-4=3a+10 equation:

a/5a-4=3a+10

We move all terms to the left:

a/5a-4-(3a+10)=0


Domain of the equation: 5a!=0

a!=0/5

a!=0

a∈R


We get rid of parentheses

a/5a-3a-10-4=0

We multiply all the terms by the denominator


a-3a*5a-10*5a-4*5a=0

Wy multiply elements

-15a^2+a-50a-20a=0

We add all the numbers together, and all the variables

-15a^2-69a=0

a = -15; b = -69; c = 0;
Δ = b2-4ac
Δ = -692-4·(-15)·0
Δ = 4761
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4761}=69$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-69)-69}{2*-15}=\frac{0}{-30}
=0
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-69)+69}{2*-15}=\frac{138}{-30}
=-4+3/5
$