2r^2+12r-14=0

Solution for 2r^2+12r-14=0 equation:

2r^2+12r-14=0

a = 2; b = 12; c = -14;
Δ = b2-4ac
Δ = 122-4·2·(-14)
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-16}{2*2}=\frac{-28}{4}
=-7
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+16}{2*2}=\frac{4}{4}
=1
$