How do you solve #2x^2 - 15x = 8#?

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Answer 1 · 2017-01-31T14:42:14

The solutions are #S={-1/2, 8}#

Let's rewrite the equation

#2x^2-15x-8=0#

And compare this to the quadratic equation

#ax^2+bx+c=0#

We start by calculating the discriminant

#Delta=b^2-4ac=(-15)^2-4(2)(-8)=225+64=289#

The solutions are

#x=(-b+-sqrtDelta)/(2a)#

#x_1=(15+sqrt289)/(2*2)=(15+17)/4=8#

#x_2=(15-17)/(2*2)=-2/4=-1/2#