Salve the following quadratic equations by factorization method. `x^2+2sqrt(2)x-6=0` `sqrt(3)x^2+10 x=7sqrt(3)=0`

Explanation

Given equation with `ax^(2)+bx+c=0` we get
On comparing with `ax^(2)+bx+c=0` we get
`a=1,b=2sqrt(2)` and `c=-6`
By quadratic formula `x=(-b+-sqrt(b^(2)-4ac))/(2a)`
`(-(2sqrt(2))+-sqrt((2sqrt(2))^(2)-4(1)(-6)))/(2(1))=(-2sqrt(2)+-sqrt(8+24))/2`
`=(-2sqrt(2)+-sqrt(32))/2,(-2sqrt(2)+-4sqrt(2))/2`
`=(-2sqrt(2)+4sqrt(2))/2,(-2sqrt(2)-4sqrt(2))/2=sqrt(2),-3sqrt(2)`
So `sqrt(2)` and `-3sqrt(2)` are the roots of the given equation.