(1/4)a-12+(7/4)a=4

Solution for (1/4)a-12+(7/4)a=4 equation:

(1/4)a-12+(7/4)a=4

We move all terms to the left:

(1/4)a-12+(7/4)a-(4)=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

(+1/4)a+(+7/4)a-12-4=0

We add all the numbers together, and all the variables

(+1/4)a+(+7/4)a-16=0

We multiply parentheses

a^2+7a^2-16=0

We add all the numbers together, and all the variables

8a^2-16=0

a = 8; b = 0; c = -16;
Δ = b2-4ac
Δ = 02-4·8·(-16)
Δ = 512
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:

$\sqrt{\Delta}=\sqrt{512}=\sqrt{256*2}=\sqrt{256}*\sqrt{2}=16\sqrt{2}$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-16\sqrt{2}}{2*8}=\frac{0-16\sqrt{2}}{16}
=-\frac{16\sqrt{2}}{16}
=-\sqrt{2}
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+16\sqrt{2}}{2*8}=\frac{0+16\sqrt{2}}{16}
=\frac{16\sqrt{2}}{16}
=\sqrt{2}
$