10r^2+31r-14=0

Solution for 10r^2+31r-14=0 equation:

10r^2+31r-14=0

a = 10; b = 31; c = -14;
Δ = b2-4ac
Δ = 312-4·10·(-14)
Δ = 1521
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1521}=39$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(31)-39}{2*10}=\frac{-70}{20}
=-3+1/2
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(31)+39}{2*10}=\frac{8}{20}
=2/5
$