5r^2-4r-5=0

Solution for 5r^2-4r-5=0 equation:

5r^2-4r-5=0

a = 5; b = -4; c = -5;
Δ = b2-4ac
Δ = -42-4·5·(-5)
Δ = 116
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:

$\sqrt{\Delta}=\sqrt{116}=\sqrt{4*29}=\sqrt{4}*\sqrt{29}=2\sqrt{29}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-2\sqrt{29}}{2*5}=\frac{4-2\sqrt{29}}{10}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+2\sqrt{29}}{2*5}=\frac{4+2\sqrt{29}}{10}
$