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Quadratic Formula answers help?????

0 votes

 

2n^2 - 7n -30 = 0
2r^2 +2r - 40 = 0

asked Mar 15, 2013 in PRE-ALGEBRA by mathgirl Apprentice

3 Answers

0 votes

2n2 -7n -30 = 0

2n2- 12n + 5n -30 = 0

Take out common factor 2n and 5

2n(n - 6) + 5(n - 6) = 0

Take out common factor n - 6

(n - 6)(2n + 5) = 0

ab = 0 then a =0 or b = 0

n - 6 = 0 or 2n + 5 = 0

n - 6 = 0

Add 6 to each side

n - 6  + 6 = 6

n = 6

2n + 5 = 0

Subtract 5 from each side

2n +5 - 5 = -5

2n = -5

Multiply each side by 2

n = -5 / 2

The roots of the equation : n = 6 or -5 / 2.

2r2+ 2r  - 40 = 0

2r2 +10r -8r  -40 = 0

Take out common factor 2r and -8

2r(r + 5) -8(r + 5) = 0

Take out common factor  r + 5

(r + 5)(2r - 8) = 0

ab = 0 then a =0 or b = 0

r +5 = 0 or 2r - 8 =0

r+ 5 =0

Subtract 5 from each side

r = -5

2r - 8 = 0

Add 8 to each side

2r = 8

Divide each side by 2

r = 4

The roots of the equation -5 and 4.

 

 

 

answered Mar 16, 2013 by diane Scholar
0 votes

 

2n^2 - 7n -30 = 0
2n^2 - 12n + 5n -30 = 0
2n (n - 6) + 5 (n - 6) = 0
(2n + 5) (n -6) = 0
2n + 5 =0 or n -6 =0
n = -5/2 or n = 6

answered Mar 17, 2013 by John Lyn Pupil
0 votes

Solve by quadratic formula.

1) The equation

image

Compare it to quadratic formula

image

image

Roots are

image

image

Discriminant is positive, so there are two real roots.

image

image

image

Solution

image

2) The equation

image

Compare it to quadratic formula

image

Roots are

image

image

image

Discriminant is positive, so there are two real roots.

image

image

image

Solution

image

answered Jun 30, 2014 by david Expert

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