Use the quadratic formal to solve the equation (show all work)

(m-4)(3m-10)=3(m-2)+18

Use the quadratic formal to solve the equation (show all work)

(m-4)(3m-10)=3(m-2)+18

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Aliquippa, PA

(m-4)(3m-10)=3(m-2)+18

3m^{2} -22m + 40 = 3m + 12

3m^{2} - 25m + 28 = 0

which is in the form ax^{2} + bx +c = 0

In this case a = 3; b = -25 and c = 28

m = [-b ± (b^{2} - 4ac)^{½}]/2a

m = {25 ± [(-25)^{2 }- 4(3)(28)]^{½}}/(2)(3)

m = {25 ± [625 - 336]^{½}}/6

m = [25 ± 17]/6

Does (7 - 4)[(3(7) - 10] = 3(7 - 2) + 18?

(3)(11) = 3(5) + 18?

Why don't you try substituting m = 1.3333 back into the original equations to see if it also "works."

Wilton, CT

(m-4)(3m-10)=3(m-2)+18 expand to get a quadratic equation as

3m²-22m+40=3m-6+18

3m²-25m+28=0 which, by quadratic formula, gives

m=(-(-25)±√(625-4(3)(28)))/6

m=(25±√(625-336))/6=(25±√289)/6=(25±17)/6=7 and 8/6 or 7, and 4/3

Middletown, CT

Hi Again Theresa;

(m-4)(3m-10)=3(m-2)+18

Let's convert this into a quadratic equation...

Let's foil the equation on the left side...

(m-4)(3m-10)

FIRST...3m^{2}

OUTER...-10m

INNER...-12m

LAST...40

3m^{2}-22m+40=3(m-2)+18

3m^{2}-22m+40=3m-6+18=3m+12

Let's subtract 12 from both sides...

-12+3m^{2}-22m+40=3m+12-12

3m^{2}-22m+28=3m

Let's subtract 3m from both sides...

-3m+3m^{2}-22m+28=3m-3m

3m^{2}-25m+28=0

Let's factor...

We know that...

FIRST...must be (3m)(m)

INNER and OUTER....-3(?)m+-(??)m

LAST....(-7)(-4) or (-14)(-2)

(3m-4)(m-7)=0

Either

3m-4=0

or m-7=0

m=4/3

or

m=7

Let's check our work...

The original equation was...

(m-4)(3m-10)=3(m-2)+18

m=4/3

[(4/3)-4][3(4/3)-10]=3[(4/3)-2]+18

[-8/3][4-10]=3(-2/3)+18

[-8/3][-6]=-2+18

16=16

(m-4)(3m-10)=3(m-2)+18

m=7

(7-4)[3(7)-10]=3(7-2)+18

3(21-10)=3(5)+18

33=33

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