(4p^2)-28p+49=169

Solution for (4p^2)-28p+49=169 equation:

(4p^2)-28p+49=169

We move all terms to the left:

(4p^2)-28p+49-(169)=0

We add all the numbers together, and all the variables

4p^2-28p-120=0

a = 4; b = -28; c = -120;
Δ = b2-4ac
Δ = -282-4·4·(-120)
Δ = 2704
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2704}=52$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-28)-52}{2*4}=\frac{-24}{8}
=-3
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-28)+52}{2*4}=\frac{80}{8}
=10
$