14r^2-19r-40=0

Solution for 14r^2-19r-40=0 equation:

14r^2-19r-40=0

a = 14; b = -19; c = -40;
Δ = b2-4ac
Δ = -192-4·14·(-40)
Δ = 2601
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2601}=51$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-19)-51}{2*14}=\frac{-32}{28}
=-1+1/7
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+51}{2*14}=\frac{70}{28}
=2+1/2
$