# If there's a problem like 3x^2 - 4x - 7 = 0, can that be solved the same way as the quadratic equation formula?

Can it be solved the same way as the quadratic equation formula ax^2 + bx + c = 0? (Except obviously instead of the + there's - )

### 5 Answers

- llafferLv 71 month agoFavourite answer
Yes. You just use negative values for b and c:

a = 3

b = -4

c = -7

x = [ -b ± √(b² - 4ac)] / (2a)

x = [ -(-4) ± √((-4)² - 4(3)(-7))] / (2 * 3)

x = [ 4 ± √(16 + 84)] / 6

x = [ 4 ± √(100)] / 6

x = (4 ± 10) / 6

x = -6/6 and 14/6

x = -1 and 7/3

- PinkgreenLv 71 month ago
3x^2-4x-7=0

=>

(3x-7)(x+1)=0

=>

x=7/3 or x=-1

Using the quadratic formula, get

a=3

b=-4

c=-7

=>

x=[4+/-sqr((-4)^2-4*3(-7)]/(2*3)

=>

x=[4+/-sqr(16+84]/6

=>

x=[4+/-sqr(100)]/6

=>

x=[4+/-10]/6

=>

x=14/6=7/3

or

x=-6/6=-1

same as before.

- DavidLv 71 month ago
Yes, by using the quadratic equation formula 3x^2 -4x -7 = 0 the value of x = -1 or 7/3

- ted sLv 71 month ago
COMMENT : you really have 3 x² + (-4) x + ( - 7 ) = 0...I may already told you that subtraction and division are not operations on the real numbers...subtraction of W is the addition of the opposite , + ( - W) , and division by K is multiplication by a ' multiplicative inverse ' , A / K = A B where KB = 1

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- rotchmLv 71 month ago
Yes. Ax² + Bx + C = 0 can be solved with the quadratic eqs, no matter the A,B & C (as long A <> 0).