1. Find the roots of Quadratic Equation `x^2+10x-56=0` by the method of perfect square

Solution:
`x^2+10x-56=0`

`=>x^2+10x-56 = 0`

Comparing the given equation with the standard quadratic equation `ax^2+bx+c=0,`

we get, `a=1, b=10, c=-56.`

`:. Delta=b^2-4ac`

`=(10)^2-4 (1) (-56)`

`=100+224`

`=324`

`:. sqrt(Delta)=sqrt(324)=18`

Now, `alpha=(-b+sqrt(Delta))/(2a)`

`=(-(10)+18)/(2*1)`

`=8/2`

`=4`

and, `beta=(-b-sqrt(Delta))/(2a)`

`=(-(10)-18)/(2*1)`

`=-28/2`

`=-14`