(3/4)a+12=42

Solution for (3/4)a+12=42 equation:

(3/4)a+12=42

We move all terms to the left:

(3/4)a+12-(42)=0


Domain of the equation: 4)a!=0

a!=0/1

a!=0

a∈R


We add all the numbers together, and all the variables

(+3/4)a+12-42=0

We add all the numbers together, and all the variables

(+3/4)a-30=0

We multiply parentheses

3a^2-30=0

a = 3; b = 0; c = -30;
Δ = b2-4ac
Δ = 02-4·3·(-30)
Δ = 360
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:

$\sqrt{\Delta}=\sqrt{360}=\sqrt{36*10}=\sqrt{36}*\sqrt{10}=6\sqrt{10}$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6\sqrt{10}}{2*3}=\frac{0-6\sqrt{10}}{6}
=-\frac{6\sqrt{10}}{6}
=-\sqrt{10}
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6\sqrt{10}}{2*3}=\frac{0+6\sqrt{10}}{6}
=\frac{6\sqrt{10}}{6}
=\sqrt{10}
$