How do you solve 4v2−3v=15−v?

2 Answers
Jarob R G.
May 3, 2018

v=−5,4

Explanation:

We start by cross-multiplying to get rid of the fractions.

4v2−3v=15−v→4(5−v)=v2−3v

This can be rearranged:

4(5−v)=v2−3v→20−4v=v2−3v→v2+v−20=0
→(v+5)(v−4)=0→v=−5,4.

We've found two solutions, v=−5 and v=4.

Phiri
May 3, 2018

v=4or−5

Explanation:

You'll want to start by cross multiplying.

4(5−v)=1(v2−3v)
20−4v=v2−3v

Have like terms on the same side.
20=v2+v

Now you can figure out in your head that v=4, but if you use the quadratic formula, you'll get another possible solution.

v2+v−20=0

v=−b±√b2−4ac2a

Plug in your values (a = 1, b = 1, c = -20), and you get:

v=−1±√12−(4⋅1⋅−20)2⋅1

Now let's simplify.

v=−1±√1+802

v=−1±√812

v=−1±92

−1+92=82=4

−1−92=82=−5

So v can be either 4 or −5

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