21r2+29r-10=0

Solution for 21r2+29r-10=0 equation:

21r^2+29r-10=0

a = 21; b = 29; c = -10;
Δ = b2-4ac
Δ = 292-4·21·(-10)
Δ = 1681
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1681}=41$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(29)-41}{2*21}=\frac{-70}{42}
=-1+2/3
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(29)+41}{2*21}=\frac{12}{42}
=2/7
$