-2(3r-14)=4(13-r)r=

Solution for -2(3r-14)=4(13-r)r= equation:

-2(3r-14)=4(13-r)r=

We move all terms to the left:

-2(3r-14)-(4(13-r)r)=0

We add all the numbers together, and all the variables

-2(3r-14)-(4(-1r+13)r)=0

We multiply parentheses

-6r-(4(-1r+13)r)+28=0


We calculate terms in parentheses: -(4(-1r+13)r), so:

4(-1r+13)r

We multiply parentheses

-4r^2+52r

Back to the equation:

-(-4r^2+52r)


We get rid of parentheses

4r^2-52r-6r+28=0

We add all the numbers together, and all the variables

4r^2-58r+28=0

a = 4; b = -58; c = +28;
Δ = b2-4ac
Δ = -582-4·4·28
Δ = 2916
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2916}=54$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-58)-54}{2*4}=\frac{4}{8}
=1/2
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-58)+54}{2*4}=\frac{112}{8}
=14
$