-4(r2)=4(2-4r)

Solution for -4(r2)=4(2-4r) equation:

-4(r2)=4(2-4r)

We move all terms to the left:

-4(r2)-(4(2-4r))=0

We add all the numbers together, and all the variables

-4r2-(4(-4r+2))=0

We add all the numbers together, and all the variables

-4r^2-(4(-4r+2))=0


We calculate terms in parentheses: -(4(-4r+2)), so:

4(-4r+2)

We multiply parentheses

-16r+8

Back to the equation:

-(-16r+8)


We get rid of parentheses

-4r^2+16r-8=0

a = -4; b = 16; c = -8;
Δ = b2-4ac
Δ = 162-4·(-4)·(-8)
Δ = 128
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:

$\sqrt{\Delta}=\sqrt{128}=\sqrt{64*2}=\sqrt{64}*\sqrt{2}=8\sqrt{2}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-8\sqrt{2}}{2*-4}=\frac{-16-8\sqrt{2}}{-8}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+8\sqrt{2}}{2*-4}=\frac{-16+8\sqrt{2}}{-8}
$