7a^2+51a+14=0

Solution for 7a^2+51a+14=0 equation:

7a^2+51a+14=0

a = 7; b = 51; c = +14;
Δ = b2-4ac
Δ = 512-4·7·14
Δ = 2209
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2209}=47$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(51)-47}{2*7}=\frac{-98}{14}
=-7
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(51)+47}{2*7}=\frac{-4}{14}
=-2/7
$