p^2+144p+2025=0

Solution for p^2+144p+2025=0 equation:

p^2+144p+2025=0

a = 1; b = 144; c = +2025;
Δ = b2-4ac
Δ = 1442-4·1·2025
Δ = 12636
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{12636}=\sqrt{324*39}=\sqrt{324}*\sqrt{39}=18\sqrt{39}$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(144)-18\sqrt{39}}{2*1}=\frac{-144-18\sqrt{39}}{2}
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(144)+18\sqrt{39}}{2*1}=\frac{-144+18\sqrt{39}}{2}
$