-1/4r-12=r+3

Solution for -1/4r-12=r+3 equation:

-1/4r-12=r+3

We move all terms to the left:

-1/4r-12-(r+3)=0


Domain of the equation: 4r!=0

r!=0/4

r!=0

r∈R


We get rid of parentheses

-1/4r-r-3-12=0

We multiply all the terms by the denominator


-r*4r-3*4r-12*4r-1=0

Wy multiply elements

-4r^2-12r-48r-1=0

We add all the numbers together, and all the variables

-4r^2-60r-1=0

a = -4; b = -60; c = -1;
Δ = b2-4ac
Δ = -602-4·(-4)·(-1)
Δ = 3584
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{3584}=\sqrt{256*14}=\sqrt{256}*\sqrt{14}=16\sqrt{14}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-60)-16\sqrt{14}}{2*-4}=\frac{60-16\sqrt{14}}{-8}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-60)+16\sqrt{14}}{2*-4}=\frac{60+16\sqrt{14}}{-8}
$