find the zeros of each equation?

via completing the square:

2x^2+8x-30=x+3

step on rearrange so as the 'x's are on one side and the constants are the other

=> 2x^2 + 7x = 33

now that divide both sides by 2 to get the coefficient of the x^2 term to equal 1

=> x^2 + (7/2)x = 33/2

now square half of the coefficient of the x^1 term and add it to both sides

=> x^2 + (7/2)x + (7/4)^2 = 33/2 + (7/4)^2

you now have the pattern of x^2 + bx + (b/2)^2 = (x + b/2)^2

=> (x + 7/4)^2 = 313/16

now take the square root of both sides

=> x + 7/4 = ±√(313)/4

now isolate the x

=> x = (-7-√313)/4 or (√313 - 7)/4

ok i so i will find it for you

there is one zero in 30 which is 0 and thats it!

easy