Please help me solve this quadratic equation?

3d^2-2d=8

2 Answers
May 3, 2018

3d^2-2d-8=03d22d8=0

d=(-(-2)+-sqrt((-2)^2-4*3*(-8)))/(2*3)d=(2)±(2)243(8)23

d=(2+-sqrt(100))/(6)d=2±1006

d=(2+-10)/(6)d=2±106

d=(2+10)/(6)=(12)/(6)=2d=2+106=126=2

d=(2-10)/(6)=(-8)/(6)=-8/6=-4/3d=2106=86=86=43

Explanation:

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

3d^2-2d-8=03d22d8=0

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

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

x=(-b+-sqrt(b^2-4ac))/(2a)x=b±b24ac2a

Which will look like,

d=(-(-2)+-sqrt((-2)^2-4*3*(-8)))/(2*3)d=(2)±(2)243(8)23

I have replaced xx here with dd, because that's what the assignment is looking for by the way.

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

d=2d=2 and d=-4/3d=43

May 3, 2018

d=2 or d=-4/3d=2ord=43

Explanation:

Here,

3d^2-2d=83d22d=8

3d^2-2d-8=03d22d8=0

Comparing with ax^2+bx+c=0ax2+bx+c=0,we get

a=3,b=-2,c=-8and xtoda=3,b=2,c=8andxd

triangle=b^2-4ac=4-4(3)(-8)=4+96=100=b24ac=44(3)(8)=4+96=100

sqrt(triangle)=10=10

So,

d=(-b+-sqrt(triangle))/(2a)=(2+-10)/(2xx3)=(2(1+-5))/(2xx3)= (1+-5)/3d=b±2a=2±102×3=2(1±5)2×3=1±53

=>d=(1+5)/3 or d=(1-5)/3d=1+53ord=153

=>d=2 or d=-4/3d=2ord=43