Problem # 1 w2 – 2w = 15

w2 - 2w -15 = 15-15

w2- 2w -15 = 0

(w – 5) (w+3) = 0

w-5 = 0 w +3 = 0 w-5 + 5 = 5+ 0 w +3-3 = 0 - 3 w = 5 w = -3

w = 5, -3

(5)2 – 2(5) = 15
25 – 10 = 15
15 = 15 True

(-3)2 – 2 (-3) = 15
9 + 6 = 15
15 =15

This is the original quadratic equation. In order to solve it, we will set it equal to zero.

Subtract 15 from both sides of the equal sign

Now that our trinomial is equal to zero we can start factoring the equation. In order to do that we have to factor the trinomial into two binomials because we cannot use the completing the square method.

The reason why I picked -5 and 3 is because when we multiply both numbers we get -15 and when we add them they equal -2.

Now we will find the value of the variables by isolating them. Solution

Let’s check our first solution (5) by plugging it in to our original quadratic equation.

Our second solution also equals to 15.

Problem #82
3v2 + 4v -1 = 0

x= -b±b2-4ac2a

a = 3 b =4 c =-1

v = -4±42-4(3)(-1)2(3)

The second quadratic equation will be solved using the quadratic formula.

Quadratic Formula

These three variables will be plugged into our formula.

Now we will multiply the given values

v = -4±286

v = -4±(4)(7)6 27

v = -4±276 Divide by 2 (numerator and denominator)

v = --2±73

v = {.215 , -1.549}

After simplifying the equation we can see that our discriminant b2 – 4ac is equal to 16 + 12 or 28. This means that our quadratic equation will have two solutions.

Our next step will be simplify our radical.

Our next step will be to simplify our equation and divide the rest of the quadratic formula by 2 in order to simplify it one more time.

Approximately