(1/4)(16+12x)=28

Solution for (1/4)(16+12x)=28 equation:

(1/4)(16+12x)=28

We move all terms to the left:

(1/4)(16+12x)-(28)=0


Domain of the equation: 4)(16+12x)!=0

x∈R


We add all the numbers together, and all the variables

(+1/4)(12x+16)-28=0

We multiply parentheses ..

(+12x^2+1/4*16)-28=0

We multiply all the terms by the denominator


(+12x^2+1-28*4*16)=0

We get rid of parentheses

12x^2+1-28*4*16=0

We add all the numbers together, and all the variables

12x^2-1791=0

a = 12; b = 0; c = -1791;
Δ = b2-4ac
Δ = 02-4·12·(-1791)
Δ = 85968
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{85968}=\sqrt{144*597}=\sqrt{144}*\sqrt{597}=12\sqrt{597}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-12\sqrt{597}}{2*12}=\frac{0-12\sqrt{597}}{24}
=-\frac{12\sqrt{597}}{24}
=-\frac{\sqrt{597}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+12\sqrt{597}}{2*12}=\frac{0+12\sqrt{597}}{24}
=\frac{12\sqrt{597}}{24}
=\frac{\sqrt{597}}{2}
$