How do I use the quadratic formula (ax^2+bx+c=0) to solve this equation?

Question

2x^2 + 9x - 5 = 0
3g^2 - 16g + 5 = 0
20a^2 + 13a + 2 = 0

Answer 1

No, the quadratic formula is

x = [-b +/- SQRT(b^2 - 4ac)] / 2a

When you have a quadratic in the form of

ax^2+bx+c=0,

you can just plug in a, b, and c into the quadratic formula to solve for x.

For example, in your third formula, "a" is the "x" you need to solve for, and you would sub in 20 for a, 13 for b, and 2 for c

a = [-13 +/- SQRT(13^2 - 4(20)(2)] / 2(20) ==

a = [-13 +/- SQRT(169 - 160)] / 40 ==

a = [-13 +/- SQRT(9)] / 40 ==

a = [ -13 +/- 3] / 40 ==

a = -16/40 or -10/40 ==

a = -2/5, -1/4

Answer 2

the discriminant surely supplies a scale of ways some distance the recommendations are aside. to illustrate the quadratic with recommendations at x=3 and x=5 has equation x^2 -8x+15=0. The discriminant is sixty 4-60 = 4, the sq. of the area between the recommendations. If the 1st term (A) isn't one, you ought to alter to illustrate the equation 3x^2 -24x+40 5=0 additionally has recommendations at x=3 and x=5. The discriminant right this is 36, it particularly is (A * distance)^2... in short, while you're taking the sq. root of the discriminant and divide with the help of A, you get the area between the two recommendations. If there are no genuine recommendations, then the area is between 2 imaginary numbers... ok, waiting for this, there is likewise a "discriminant" for cubic equations... ask your instructor to tell you that one...

Answer 3

2x^2 + 9x - 5 = 0
a = 2, b - 9, c = -5
x1 = [-b + sqrt.(b^2 - 4ac)] / 2a

= [ - 9 + sqrt(81 + 40)] / 4

= (- 9 + 11) / 4 = 2/4 = 1/2

x2 = [-b - sqrt(b^2 - 4ac)] / 2a

= [ - 9 - sqrt. 121] / 4

= [ - 9 - 11] / 4
= -20 / 4 = -5

Try the other equations in a similar way. Best wishes.

Answer 4

x=[-b±√(b²-4ac)] /(2a)]
1) a= 2 b=9 c =-5 by subistitution x= 0.5 or x =-5
2) a= 3 b = -16 c=5 '' g=5 or g =1/3 =0.3333
3)a=20 b = 13 c=2 '' a= -0.25 or a =-0.4

Answer 5

I think you have the wrong quadratic formula.

.... -b±sqrt(b^2-4ac)
x = -----------------------
..............2a

first one would be
(-9±sqrt(9^2-4*2*-5)) / (2*2) = (-9±sqrt(81+40)) / 4 =
=( -9±11)/4 = -5 and +1/2

The others are done the same way.
Answers are 1/3 and 5 for second eq. and -2/5 and -1/4 for the third.

Answer 6

identify the a,b & c subsitute into the formula and you will get x

refer to the reference
1) x=0.5 ,-5 (a=2,b=9,c=(-5))
2) x=5,0.3333 (a=3,b=(-16),c=(5))
3) x=-0.25,-0.4 (a=20,b=(13),c=(2))