3x^2-4x-5=0 How do I solve this using the quadratic formula?

Answer 1

-(-4)±√(-4)²-4(3)(-5) over 2(3)



4±√76 over 6


simplified that is

2±√76 over 3


and depending how you need the answer it could be simplified even more as

2±2√19 over 3


this is because √76 is 4√19
hope this makes sense

Answer 2

The Quadratic formula= b + or - sqr root of (b^2-4(a*c)
all divided by 2a

3x^2-4x-5 =

A= 3
B=-4
C=-5

Plug in and solve

x1 = 2.1196
x2 = -.7863

Remember to do it for b + and b- the square root of the rest of the equations.

Answer 3

The general quad formula is

ax^2 + bx + c

The quadratic formula solution is
x = (-b +- (b^2 - 4ac)^0.5)/2a

For your formula a = 3, b= - 4, c= -5

Plugging into the formula

x = (-(-4) +- (16 - 4(3)(-5))^0.5)/6

x = (2 +- 2(19)^0.5) / 3

Answer 4

boring?? This is my idea of a good time.. (shoot me, please..)

x=((-b)+-sqrt(b^2-4ac))/2a

a=3 b=-4 and c=-5, now plug in and solve...

x=(4+-sqrt(16+(-4)(3)(-5)))/6
x=(4+-sqrt(16+60))/6
x=(4+-sqrt(76))/6
x=(4+2sqrt(19))/6 AND x=(4-2sqrt(19))/6
x=(2+sqrt(19))/3 AND x=(2-sqrt(19))/3

This is simplest form.. if you substitute 4.359 for sqrt(19) you can solve farther

x=(2+4.359)/3 AND x=(2-4.359)/3
x=2.120 AND x=-.786

Answer 5

don't be boring ..........it's yahoo....enjoy......leave all maths to your school..........