What are the zeros of the quadratic function f(x)=6x^2+12x-7

Use the quadratic formula

[-12 +/- square root (12^2- (4)(6)(-7))]/2(6)

[-12 +/- square root (312)]/12

-1+(1/6) square root (78)

and 

-1-(1/6) square root (78)

intensity of the vector is v= √37 ≈ 6.08 units and make angle ∡α ≈ 9.46° with east direction

step-by-step explanation:

required vector is consists of the two componentsvx= 2+4=6 units   and vy= 1 unit and vx ⊥ vywe will use pythagorian theorem to find intensity of the vector vv∧2 = vx∧2 + vy∧2 => v = √vx∧2 + vy∧2 = √6∧2 + 1∧2 = √36+1 = √37 ≈ 6.08 unitsthe angle ∡α between vector and east direction we wil find with tanαtanα = 1/6 => α = arc tanα = arc 1/6 => α ≈ 9.46°

hope this

$8

step-by-step explanation:

it is pretty much backwards.

you take the 4x2 multiply that, and that is = $8