r^2-4r-59=0

Solution for r^2-4r-59=0 equation:

r^2-4r-59=0

a = 1; b = -4; c = -59;
Δ = b2-4ac
Δ = -42-4·1·(-59)
Δ = 252
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:

$\sqrt{\Delta}=\sqrt{252}=\sqrt{36*7}=\sqrt{36}*\sqrt{7}=6\sqrt{7}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-6\sqrt{7}}{2*1}=\frac{4-6\sqrt{7}}{2}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+6\sqrt{7}}{2*1}=\frac{4+6\sqrt{7}}{2}
$