(6r-7)10=1/4r

Solution for (6r-7)10=1/4r equation:

(6r-7)10=1/4r

We move all terms to the left:

(6r-7)10-(1/4r)=0


Domain of the equation: 4r)!=0

r!=0/1

r!=0

r∈R


We add all the numbers together, and all the variables

(6r-7)10-(+1/4r)=0

We multiply parentheses

60r-(+1/4r)-70=0

We get rid of parentheses

60r-1/4r-70=0

We multiply all the terms by the denominator


60r*4r-70*4r-1=0

Wy multiply elements

240r^2-280r-1=0

a = 240; b = -280; c = -1;
Δ = b2-4ac
Δ = -2802-4·240·(-1)
Δ = 79360
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:

$\sqrt{\Delta}=\sqrt{79360}=\sqrt{256*310}=\sqrt{256}*\sqrt{310}=16\sqrt{310}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-280)-16\sqrt{310}}{2*240}=\frac{280-16\sqrt{310}}{480}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-280)+16\sqrt{310}}{2*240}=\frac{280+16\sqrt{310}}{480}
$