Can someone help me with this :/ solve using the quadratic formula: x=(-b +/-sqr(b^2 - 4ac)/(2a)
2x² + x - 1 = 0

answers in (pic)

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2x² + x - 1 = 0

a= 2 b= 1 c= -1

x = [ -b +- sqr rt( b^2 -4ac) ] / 2*a

x = -1 +- sqr rt (1 - 4 *2*-1) / 4

x1 = [ -1 + (sqr rt (1 +8)) / 4

x1 = ( -1 + sqr rt (9)) / 4

x1 = (-1 + 3) / 4

x1 = .5

x2 = (-1 -3) / 4

x2 = -4 / 4

x2 = -1

And to check that http://www.1728.org/quadratc.htm

Step-by-step explanation:

so far i think you have the right thing going for you just remember soh cah toh

step-by-step explanation:

sine over hypotenuse

cosine   equals adjacent

tangent over hypotenuse