2x^2 - 5x =6 now i know i need to find the root first.........right........using the quadratic formula?

Question

please help me with these problems.............3x^2-2x+1=0........i just can't do this..........

please forgive me, but it should say =3 instead of 6........i am just trying to pass this class.........i hate this class........

would i do 3x2-2x+1=0 the same way.........find the root....for this quadratic formula....

2x^2-5x=3 solve with the quadratic formula..........please help.........

2x^2-5x=3...........what do i do help please.........................

Answer 1

2x^2 -5x - 6 =0

the roots
r1 = (- b + SQRT (b^2 - 4ac) )/2a
r2 = (- b - SQRT (b^2 - 4ac) )/2a
where
a = 2 , b= -5 , c = -6

so r1=(5+ sqrt (25+48)/4) = 3.386
r2 = (5-sqrt (25+48)/4) = -0.886

By the way, the answer just above mine has an error, the √-13 should be √73 cos it's actually √ 5^2 - 4 (2)(-6) = √25 + 48 = √73

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AGAIN, YOU JUST NEED TO CHANGE THE COEFFICIENT C
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after your details added , you want to solve
2x^2 -5x - 3 =0

the roots
r1 = (- b + SQRT (b^2 - 4ac) )/2a
r2 = (- b - SQRT (b^2 - 4ac) )/2a
where
a = 2 , b= -5 , c = -3

so r1=(5+ sqrt (25-(4)(2)(-3))/4) = (5+sqrt(49))/4 = 12/4 = 3
r2 = (5-sqrt (25-(4)(2)(-3))/4) = (5-sqrt(49))/4 = -2/4 = -1/2

Answer 2

Either the quadratic equation or a good slide rule.

Just set the left C index over the 3 on the D scale. Slide the cursor along subtracting the D value from the CI value. (The CI value is negative - if it were positive, you'd add).

At 3.3 on the D, you have .9 on the CI, which is to small (since I divided everything by 2, I'm looking for 2.5)

At 3.4, I have less than .9, which is too big.

At 3.38, I have .888 on the CI - just slightly too small at 2.492, so I have to go a little more to the right.

At 3.39 on the D, I have .885 on the CI, which is just slightly too big at 2.505.

That puts the answers at just a little over 3.385 and around .886 (a little under .8865, which half between my last two answers).

Have to remember to put the negative sign in, making your final answers around (3.386 or 3.387) and (-.886)

In other words, 3.39 and -.89.

Answer 3

this is in the form
ax^2 + bx +c=0
to get the root there is a formula
where a=2 b=5 c=6

x= -b+√ b^2-4ac divided by 2a (to get the positive root)

y= -b-√ b^2-4ac divided by 2a (to get the negetive root)

Note: x is the positive root and y is the negetive root
after i calculated i got that
the positive root is -5+√ -13 divided by 4

the negetive root is -5-√ -13 divided by 4

if u want u can simplify...
Regards,
Govi,
india

Answer 4

You should find delta ! and the formula is : b^2-4ac
and for this delta is 73>0 so it has 2 roots ! and the formula for the roots are : [-b+(delta)^(1/2)]/2a and [-b-(delta)^(1/2)]/2a

Answer 5

If you meant to type

2x^2 - 5x + 6
x = (-b ± sqrt(b^2 - 4ac))/2a

x = (-(-5) ± sqrt((-5)^2 - 4(2)(6)))/(2(2))
x = (5 ± sqrt(25 - 48))/4
x = (5 ± sqrt(-23))/4
x = (5 ± isqrt(23))/4
x = (1/4)(5 ± isqrt(23))

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but if you did mean 2x^2 - 5x = 6, then
2x^2 - 5x - 6 = 0
x = (-(-5) ± sqrt((-5)^2 - 4(2)(-6)))/(2(2))
x = (5 ± sqrt(25 + 48))/4
x = (5 ± sqrt(73))/4
x = (1/4)(5 ± sqrt(73))

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3x^2 - 2x +1
x = (-(-2) ± sqrt((-2)^2 - 4(3)(1)))/(2(3))
x = (2 ± sqrt(4 - 12))/6
x = (2 ± sqrt(-8))/6
x = (2 ± sqrt(-4 * 2))/6
x = (2 ± 2isqrt(2))/6
x = (1/3)(1 ± isqrt(2))

Answer 6

2x2 - 5x = 6
2x2 - 5x - 6 = 0
Use the abc-formule
D = 25 - 4 2 (-6) = 73
x ={5 - V73}/4 or x ={5 + V73}/4

Answer 7

Use the Dharacharya formula or the discriminanT method, as u call it.

Answer 8

yeah gotta use that or a good calculator,

roots are -.886 or 3.386