21r2+22r-8=0

Solution for 21r2+22r-8=0 equation:

21r^2+22r-8=0

a = 21; b = 22; c = -8;
Δ = b2-4ac
Δ = 222-4·21·(-8)
Δ = 1156
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1156}=34$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-34}{2*21}=\frac{-56}{42}
=-1+1/3
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+34}{2*21}=\frac{12}{42}
=2/7
$