how do I solve 4x^2-8x+3=5?

Question

I can't solve this Please help!!

Answer 1

4x^2 - 8x +3 - 5 = 0
4x^2 - 8x - 2 = 0 ...... divide everything by 2
2x^2 - 4x -1 = 0

quadratic formula [-b +/- (b^2-4ac)^1/2]/2a

x= [4 + 24^(1/2)]/4
and
x= [4 - 24^(1/2)]/4

Answer 2

u can put the 5 to the left side
that will be
4x^2 - 8x -2 = 0
then the whole thing divided by 2
u can get 2x^2 - 4x - 1 = 0
then use the abc formulae

x = -b+/- {b^2-4ac}^1/2 over 2a
x= 4 +/- 8^1/2 over 4
x = 1 +/- 0.77

Answer 3

Set the equation equal to zero by subtracting 5 from both sides.
4x^2 - 8x -2 = 0

Now use the quadratic formula to solve for x.
(-b+- rt(b^2 - 4ac))/2a
where a = 4, b = -8 and c = -2

Hope this helps. Good Luck. :)

Answer 4

Simplify: 4x^2-8x-2 = 0

Divide by 4: x^2-2x-1/2 = 0

Apply quadratic formula: x = (2 +- root(4+4*1*(1/2)))/2

x = 2.22474
x = -.22474

Answer 5

4x^2-8x+3=5
-5 -5
4x^2-8x-2=0

Then you use the quadratic formula:

-b +/- the sq root of (b^2-4ac)
all of this divided by 2a

In this equation, a is 4, b is -8 and c is -2

Answer 6

set the equation equal to zero by subtarcting 5 from both sides, this will give,
4x^2-8x-2=0
now equation is of the from ax^2+bx+c=0
now use the quadratic equation, i.e.
x={-b+sqrt(b^2-4ac)}/2a and x={-b-sqrt(b^2-4ac)}/2a
a=4,b=-8,c=-2

x={-(-8)+sqrt((-8)^2-4*4*(-2)}/2*4
x=2.2247
x={-(-8)-sqrt((-8)^2-4*4*(-2)}/2*4
x=-0.2247

Answer 7

take the quadratic formula after subtracting 5 from both sides and setting it to equal 0

Answer 8

Using the quadratic formula (as mentioned above), I get:

1 +/- .5*sqrt(6)

I substituted both answers back into the original problem, and got the right solution. :-)

Answer 9

4x^2-8x-2=0
that is 2x^2-4x-1=0
for ax^2+bx+c=0 roots are [-b+sqrt(b^2-4ac)]/2a and[-b-sqrt(b^2-4ac)]/2a
so roots are 2+sqrt6 and 2-sqrt6

Answer 10

i came up with x=2