2r^2-4r-5=4

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

2r^2-4r-5=4

We move all terms to the left:

2r^2-4r-5-(4)=0

We add all the numbers together, and all the variables

2r^2-4r-9=0

a = 2; b = -4; c = -9;
Δ = b2-4ac
Δ = -42-4·2·(-9)
Δ = 88
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:

$\sqrt{\Delta}=\sqrt{88}=\sqrt{4*22}=\sqrt{4}*\sqrt{22}=2\sqrt{22}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-2\sqrt{22}}{2*2}=\frac{4-2\sqrt{22}}{4}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+2\sqrt{22}}{2*2}=\frac{4+2\sqrt{22}}{4}
$