21r^2+26r-15=0

Solution for 21r^2+26r-15=0 equation:

21r^2+26r-15=0

a = 21; b = 26; c = -15;
Δ = b2-4ac
Δ = 262-4·21·(-15)
Δ = 1936
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1936}=44$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(26)-44}{2*21}=\frac{-70}{42}
=-1+2/3
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(26)+44}{2*21}=\frac{18}{42}
=3/7
$