7p^2+105p+392=0

Solution for 7p^2+105p+392=0 equation:

7p^2+105p+392=0

a = 7; b = 105; c = +392;
Δ = b2-4ac
Δ = 1052-4·7·392
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(105)-7}{2*7}=\frac{-112}{14}
=-8
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(105)+7}{2*7}=\frac{-98}{14}
=-7
$