2r^2-4r-3=8

Solution for 2r^2-4r-3=8 equation:

2r^2-4r-3=8

We move all terms to the left:

2r^2-4r-3-(8)=0

We add all the numbers together, and all the variables

2r^2-4r-11=0

a = 2; b = -4; c = -11;
Δ = b2-4ac
Δ = -42-4·2·(-11)
Δ = 104
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:

$\sqrt{\Delta}=\sqrt{104}=\sqrt{4*26}=\sqrt{4}*\sqrt{26}=2\sqrt{26}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-2\sqrt{26}}{2*2}=\frac{4-2\sqrt{26}}{4}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+2\sqrt{26}}{2*2}=\frac{4+2\sqrt{26}}{4}
$