21r^2+13r-20=0

Solution for 21r^2+13r-20=0 equation:

21r^2+13r-20=0

a = 21; b = 13; c = -20;
Δ = b2-4ac
Δ = 132-4·21·(-20)
Δ = 1849
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1849}=43$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-43}{2*21}=\frac{-56}{42}
=-1+1/3
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+43}{2*21}=\frac{30}{42}
=5/7
$