9a^2+20a-21=0

Solution for 9a^2+20a-21=0 equation:

9a^2+20a-21=0

a = 9; b = 20; c = -21;
Δ = b2-4ac
Δ = 202-4·9·(-21)
Δ = 1156
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1156}=34$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-34}{2*9}=\frac{-54}{18}
=-3
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+34}{2*9}=\frac{14}{18}
=7/9
$