How do you solve 3y^2 - 7y + 4 = 03y2−7y+4=0 using the quadratic formula?

2 Answers
Narad T.
Jul 18, 2018

The solutions are S={1,4/3}S={1,43}

Explanation:

The quadratic equation is

3y^2-7y+4=03y2−7y+4=0

The discriminant is

Delta=b^2-4ac=(-7)^2-4*3*4=49-48=1

As Delta>0, there are 2 real roots

Therefore,

y=(-b+-sqrt(Delta))/(2a)

=(-(-7)+-sqrt1)/(2*3)

=(7+-1)/6

Therefore,

y_1=8/6=4/3

and

y_2=6/6=1

The solutions are S={1,4/3}

Harish Chandra Rajpoot
Jul 18, 2018

y=4/3, 1

Explanation:

Given quadratic equation:

3y^2-7y+4=0

Using quadratic formula,

y=\frac{-(-7)\pm\sqrt{(-7)^2-4(3)(4)}}{2(3)}

y=\frac{7\pm1}{6}

y=4/3, 1

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