4-(3/5)(3a+4)=7

Solution for 4-(3/5)(3a+4)=7 equation:

4-(3/5)(3a+4)=7

We move all terms to the left:

4-(3/5)(3a+4)-(7)=0


Domain of the equation: 5)(3a+4)!=0

a∈R


We add all the numbers together, and all the variables

-(+3/5)(3a+4)+4-7=0

We add all the numbers together, and all the variables

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

We multiply parentheses ..

-(+9a^2+3/5*4)-3=0

We multiply all the terms by the denominator


-(+9a^2+3-3*5*4)=0

We get rid of parentheses

-9a^2-3+3*5*4=0

We add all the numbers together, and all the variables

-9a^2+57=0

a = -9; b = 0; c = +57;
Δ = b2-4ac
Δ = 02-4·(-9)·57
Δ = 2052
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:

$\sqrt{\Delta}=\sqrt{2052}=\sqrt{36*57}=\sqrt{36}*\sqrt{57}=6\sqrt{57}$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6\sqrt{57}}{2*-9}=\frac{0-6\sqrt{57}}{-18}
=-\frac{6\sqrt{57}}{-18}
=-\frac{\sqrt{57}}{-3}
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6\sqrt{57}}{2*-9}=\frac{0+6\sqrt{57}}{-18}
=\frac{6\sqrt{57}}{-18}
=\frac{\sqrt{57}}{-3}
$