2r^2-16r+14=0

Solution for 2r^2-16r+14=0 equation:

2r^2-16r+14=0

a = 2; b = -16; c = +14;
Δ = b2-4ac
Δ = -162-4·2·14
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-12}{2*2}=\frac{4}{4}
=1
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+12}{2*2}=\frac{28}{4}
=7
$