akar persamaan kuadrat x2-2px+p2-q2+2qr-r2=0

x² - 2px + p²-q²+2qr-r² = 0
x1,2 = {-b ± √D} / 2a
= {-(-2p) ± √((-2p)² - 4(1)(p²-q²+2qr-r²)} / 2(1)
= {2p ± √(4p² - 4p² + 4q² - 8qr + 4r²)} / 2
= {2p ± √(4q²-8pr+4r²)} / 2
= {2p ± √(2q-2r)²} / 2
= {2p ± (2q-2r)} / 2
x1 = {2p + (2q - 2r)} / 2
= (2p + 2q - 2r) / 2
= p + q - r
x2 = {2p - (2q - 2r)} / 2
= (2p - 2q + 2r) / 2
= p - q + r
jadi akar2nya
p + q - r atau p - q + r