Question

Solve the above equation

Answer 1

hi there,
Given
2x^2 + 4x - 8 = 0

roots of a quadratic is found by using
ax^2 + bx + c = 0
x = (-b +√(b^2 - 4ac) / 2a
x = (-b -√(b^2 - 4ac) / 2a
here
a=2 , b =4 , c= -8
now
-b = -4
2a = 2*2 = 4
√(b^2 - 4ac) = √(16 - 4*2*(-8) ) = √( 16 + 64 ) = √80 = 4√5

so
x = (-b +√(b^2 - 4ac) / 2a
x = (-4 + 4√5) / 4
x = -1+√5

similarly
x = (-b -√(b^2 - 4ac) / 2a
x = (-4 - 4√5) / 4
x = -1-√5

so answer is x = -1+√5 , -1-√5

Hope it helped.
Let me know if any doubts.
Cheers !!!