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