Solve the following quadratic equations. Show the work so I can learn it?

Question

4x^2=13x+12

Answer 1

The general formula of a 2nd degrree polynomial is
ax^2 + bx + c = 0

Rewritting the equation in this form, you get 4x^2 - 13x - 12 = 0
Thus, a = 4, b = -13, c= -12

The quadratic formula is (-b[+-] sqrt(b^2-4ac))/2a

Plugging in the values, you get (13[+-]sqrt(169+192))/8

=(13 [+-] 19)/8

Thus your two answers are (13+19)/8 = 4, (13-19)/8 = -3/4

Answer 2

4x^2 = 13x + 12

4x^2 - 13x - 12 = 0

the discriminant of the quadratic expression on the left-hand side is b^2 - 4ac = (-13)^2 - 4(4)(-12) = 169 + 16(12) = 361 = (19)^2, which is a perfect square. hence, the quadratic expression is factorable as:

(x - 4)(4x + 3) = 4x^2 - 16x + 3x - 12 = 4x^2 -13x - 12

hence, (x - 4)(4x + 3) = 0.

consequently, either x = 4 or x = -3/4.

for additional information, try looking up "quadratic formula" on the InterNet.

Answer 3

you have to make the equation equal to zero:

0= 4x^2+13x+12

then put it in the quadratic formula. Sorry!

Answer 4

4x^2=13x+12,it can be written as
4x^2-13x-12=0,now
4x^2-16x+3x-12=0
4x(x-4) +3(x-4)=0
either x-4=0 or 4x+3=0
so two values are
x=4, x= -3/4

Answer 5

"4x^2=13x+12"

4x^2 - (13x+12)=13x+12 - (13x+12)
4x^2-13x-12=0

a=4; b=-13; c=-12
Use the quadratic formula.
-b+/-(sqrt)b^2-4ac/2a

Plug in...

-(-13)+/-(sqrt)-13^2-4(4)(-12) / 2(4)
13+/-(sqrt)169+192 / 8
13+/-(sqrt)361 / 8
The square root of 361 is "19"
So...
13+/-19 / 8
You will have two answers...

13+19/8
=32/8
=4

13-19/8
=-6/8
=-3/4

So your answers are... "4" and "-3/4".

YOU'RE WELCOME IN ADVANCE. IM GLAD TO HELP.

Answer 6

4x^2-13x-12=0
(4x+3)(x-4)=0

Answer 7

4x^2 = 13x + 12
4x^2 - 13x - 12 = 0

x = (-b ± sqrt(b^2 - 4ac))/(2a)

x = (-(-13) ± sqrt((-13)^2 - 4(4)(-12)))/(2(4))
x = (13 ± sqrt(169 + 192))/8
x = (13 ± sqrt(361))/8
x = (13 ± 19)/8
x = (32/8) or (-6/8)
x = 4 or (-3/4)

Answer 8

4x^2-13x-12=0
x= [b +- (b^2 - 4ac)^1/2]/2a

=>
x= 4 , -3/4