5r^2-6r-18=0

Solution for 5r^2-6r-18=0 equation:

5r^2-6r-18=0

a = 5; b = -6; c = -18;
Δ = b2-4ac
Δ = -62-4·5·(-18)
Δ = 396
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{396}=\sqrt{36*11}=\sqrt{36}*\sqrt{11}=6\sqrt{11}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-6\sqrt{11}}{2*5}=\frac{6-6\sqrt{11}}{10}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+6\sqrt{11}}{2*5}=\frac{6+6\sqrt{11}}{10}
$