r^2+14r+8=4r

Solution for r^2+14r+8=4r equation:

r^2+14r+8=4r

We move all terms to the left:

r^2+14r+8-(4r)=0

We add all the numbers together, and all the variables

r^2+10r+8=0

a = 1; b = 10; c = +8;
Δ = b2-4ac
Δ = 102-4·1·8
Δ = 68
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:

$\sqrt{\Delta}=\sqrt{68}=\sqrt{4*17}=\sqrt{4}*\sqrt{17}=2\sqrt{17}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{17}}{2*1}=\frac{-10-2\sqrt{17}}{2}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{17}}{2*1}=\frac{-10+2\sqrt{17}}{2}
$