Question

How do you solve `-2x^{2} - 14x + 5= 0`?

Answer 1

2nd degree polynomial...

Explanation:

lets find `Delta`
`Delta=b^2-4a*c` where
`a=-2`,
`b=-14`,
`c=5`

`Delta = (-14)^2-4(-2*5) = 196-(4*-10) = 196+40=236`
now lets find roots (where polynomial intersects with x axis...)

`x1 = (-b-sqrt(Delta))/(2a) = (-(-14) - sqrt(236))/(2*-2) = (14-sqrt(236))/(-4)`

`x1 = (14-15.3623)/(-4)=(-1.3623)/-4=0.3405`

`x2=(-b+sqrt(Delta))/(2a)=(14+15.3623)/(-4)=29.3623/-4=-7.3405`