Please help me solve this quadratic equation?

Question

3d^2-2d=8

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Answer 1 · 2018-05-03T11:26:47

#3d^2-2d-8=0#

#d=(-(-2)+-sqrt((-2)^2-4*3*(-8)))/(2*3)#

#d=(2+-sqrt(100))/(6)#

#d=(2+-10)/(6)#

#d=(2+10)/(6)=(12)/(6)=2#

#d=(2-10)/(6)=(-8)/(6)=-8/6=-4/3#

We can analyze the equation after we get all the numbers on one side,

#3d^2-2d-8=0#

From here, we can see that #a=3#, #b=-2# and #c=-8#.

We now need to put it into the quadratic equation formula.

#x=(-b+-sqrt(b^2-4ac))/(2a)#

Which will look like,

#d=(-(-2)+-sqrt((-2)^2-4*3*(-8)))/(2*3)#

I have replaced #x# here with #d#, because that's what the assignment is looking for by the way.

When we do the quadratic equation, we will get the answers.

#d=2# and #d=-4/3#

Answer 2 · 2018-05-03T11:39:01

#d=2 or d=-4/3#

Here,

#3d^2-2d=8#

#3d^2-2d-8=0#

Comparing with #ax^2+bx+c=0#,we get

#a=3,b=-2,c=-8and xtod#

#triangle=b^2-4ac=4-4(3)(-8)=4+96=100#

#sqrt(triangle)=10#

So,

#d=(-b+-sqrt(triangle))/(2a)=(2+-10)/(2xx3)=(2(1+-5))/(2xx3)= (1+-5)/3#

#=>d=(1+5)/3 or d=(1-5)/3#

#=>d=2 or d=-4/3#