4(2r-1)=-2r(3r+16)

Solution for 4(2r-1)=-2r(3r+16) equation:

4(2r-1)=-2r(3r+16)

We move all terms to the left:

4(2r-1)-(-2r(3r+16))=0

We multiply parentheses

8r-(-2r(3r+16))-4=0


We calculate terms in parentheses: -(-2r(3r+16)), so:

-2r(3r+16)

We multiply parentheses

-6r^2-32r

Back to the equation:

-(-6r^2-32r)


We get rid of parentheses

6r^2+32r+8r-4=0

We add all the numbers together, and all the variables

6r^2+40r-4=0

a = 6; b = 40; c = -4;
Δ = b2-4ac
Δ = 402-4·6·(-4)
Δ = 1696
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:

$\sqrt{\Delta}=\sqrt{1696}=\sqrt{16*106}=\sqrt{16}*\sqrt{106}=4\sqrt{106}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-4\sqrt{106}}{2*6}=\frac{-40-4\sqrt{106}}{12}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+4\sqrt{106}}{2*6}=\frac{-40+4\sqrt{106}}{12}
$