5r^2+14r-3=0

Solution for 5r^2+14r-3=0 equation:

5r^2+14r-3=0

a = 5; b = 14; c = -3;
Δ = b2-4ac
Δ = 142-4·5·(-3)
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-16}{2*5}=\frac{-30}{10}
=-3
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+16}{2*5}=\frac{2}{10}
=1/5
$