5r^2+17r+14=0

Solution for 5r^2+17r+14=0 equation:

5r^2+17r+14=0

a = 5; b = 17; c = +14;
Δ = b2-4ac
Δ = 172-4·5·14
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(17)-3}{2*5}=\frac{-20}{10}
=-2
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(17)+3}{2*5}=\frac{-14}{10}
=-1+2/5
$