2r^2-28r+62=0

Solution for 2r^2-28r+62=0 equation:

2r^2-28r+62=0

a = 2; b = -28; c = +62;
Δ = b2-4ac
Δ = -282-4·2·62
Δ = 288
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:

$\sqrt{\Delta}=\sqrt{288}=\sqrt{144*2}=\sqrt{144}*\sqrt{2}=12\sqrt{2}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-28)-12\sqrt{2}}{2*2}=\frac{28-12\sqrt{2}}{4}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-28)+12\sqrt{2}}{2*2}=\frac{28+12\sqrt{2}}{4}
$