21r^2+40r-21=0

Solution for 21r^2+40r-21=0 equation:

21r^2+40r-21=0

a = 21; b = 40; c = -21;
Δ = b2-4ac
Δ = 402-4·21·(-21)
Δ = 3364
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{3364}=58$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-58}{2*21}=\frac{-98}{42}
=-2+1/3
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+58}{2*21}=\frac{18}{42}
=3/7
$