7p^2+17p=-6

Solution for 7p^2+17p=-6 equation:

7p^2+17p=-6

We move all terms to the left:

7p^2+17p-(-6)=0

We add all the numbers together, and all the variables

7p^2+17p+6=0

a = 7; b = 17; c = +6;
Δ = b2-4ac
Δ = 172-4·7·6
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(17)-11}{2*7}=\frac{-28}{14}
=-2
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(17)+11}{2*7}=\frac{-6}{14}
=-3/7
$