4r^2-36r+32=0

Solution for 4r^2-36r+32=0 equation:

4r^2-36r+32=0

a = 4; b = -36; c = +32;
Δ = b2-4ac
Δ = -362-4·4·32
Δ = 784
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{784}=28$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-36)-28}{2*4}=\frac{8}{8}
=1
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-36)+28}{2*4}=\frac{64}{8}
=8
$