100r^2+200r-21=0

Solution for 100r^2+200r-21=0 equation:

100r^2+200r-21=0

a = 100; b = 200; c = -21;
Δ = b2-4ac
Δ = 2002-4·100·(-21)
Δ = 48400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{48400}=220$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(200)-220}{2*100}=\frac{-420}{200}
=-2+1/10
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(200)+220}{2*100}=\frac{20}{200}
=1/10
$