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

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

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

We move all terms to the left:

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

We multiply parentheses

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


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

2(3r+16)

We multiply parentheses

6r+32

Back to the equation:

-(6r+32)


We get rid of parentheses

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

We add all the numbers together, and all the variables

8r^2-10r-32=0

a = 8; b = -10; c = -32;
Δ = b2-4ac
Δ = -102-4·8·(-32)
Δ = 1124
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:

$\sqrt{\Delta}=\sqrt{1124}=\sqrt{4*281}=\sqrt{4}*\sqrt{281}=2\sqrt{281}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-2\sqrt{281}}{2*8}=\frac{10-2\sqrt{281}}{16}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+2\sqrt{281}}{2*8}=\frac{10+2\sqrt{281}}{16}
$