Find Integral values of x for which x^{2 }+ 19x + 46 is a perfect square?

** x ^{2}+19x+46**

**=x ^{2}+18x+81+x-35**

**=(x+9) ^{2}+(x-35)**

**for the above equation to become perfect square x-35 must be equal to zero**

**therefore x=35**

**similarly, x ^{2}+19x+46=x^{2}+20x+100-(x+54)**

**=(x+10) ^{2}-(x+54)**

**here x+54=0**

**=> x=-54**

**x ^{2}+19x+46=x^{2}+22x+121-3(x+25)**

**=(x+11) ^{2}-3(x+25)**

**here x+25=0**

**=> x=-25**

**x ^{2}+19x+46=x^{2}+16x+64+3x-18**

**=(x+8) ^{2}+3(x-6)**

**here x-6=0**

**=> x=6**