4r^-3(2r^2)=-

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

4r^-3(2r^2)=-

We move all terms to the left:

4r^-3(2r^2)-(-)=0

determiningTheFunctionDomain
-32r^2+4r^-(-)=0

We add all the numbers together, and all the variables

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

We add all the numbers together, and all the variables

-32r^2+4r=0

a = -32; b = 4; c = 0;
Δ = b2-4ac
Δ = 42-4·(-32)·0
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4}{2*-32}=\frac{-8}{-64}
=1/8
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4}{2*-32}=\frac{0}{-64}
=0
$