a+(3/4a)=21

Solution for a+(3/4a)=21 equation:

a+(3/4a)=21

We move all terms to the left:

a+(3/4a)-(21)=0


Domain of the equation: 4a)!=0

a!=0/1

a!=0

a∈R


We add all the numbers together, and all the variables

a+(+3/4a)-21=0

We get rid of parentheses

a+3/4a-21=0

We multiply all the terms by the denominator


a*4a-21*4a+3=0

Wy multiply elements

4a^2-84a+3=0

a = 4; b = -84; c = +3;
Δ = b2-4ac
Δ = -842-4·4·3
Δ = 7008
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{7008}=\sqrt{16*438}=\sqrt{16}*\sqrt{438}=4\sqrt{438}$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-84)-4\sqrt{438}}{2*4}=\frac{84-4\sqrt{438}}{8}
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-84)+4\sqrt{438}}{2*4}=\frac{84+4\sqrt{438}}{8}
$