r^2-4r-165=0

Solution for r^2-4r-165=0 equation:

r^2-4r-165=0

a = 1; b = -4; c = -165;
Δ = b2-4ac
Δ = -42-4·1·(-165)
Δ = 676
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{676}=26$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-26}{2*1}=\frac{-22}{2}
=-11
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+26}{2*1}=\frac{30}{2}
=15
$