10r^2+67r=21

Solution for 10r^2+67r=21 equation:

10r^2+67r=21

We move all terms to the left:

10r^2+67r-(21)=0

a = 10; b = 67; c = -21;
Δ = b2-4ac
Δ = 672-4·10·(-21)
Δ = 5329
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{5329}=73$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(67)-73}{2*10}=\frac{-140}{20}
=-7
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(67)+73}{2*10}=\frac{6}{20}
=3/10
$