2r^2+19r-10=0

Solution for 2r^2+19r-10=0 equation:

2r^2+19r-10=0

a = 2; b = 19; c = -10;
Δ = b2-4ac
Δ = 192-4·2·(-10)
Δ = 441
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{441}=21$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-21}{2*2}=\frac{-40}{4}
=-10
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+21}{2*2}=\frac{2}{4}
=1/2
$