Can you complete the square of this equation, 4x^2+7x=15?

Question

I am confused, are you supposed to divide by 4 on both sides? or subtract 15 from both sides?

Answer 1

Subtract 15 on both sides...

4x^2 + 7x - 15 = 0
4x^2 + 12x - 5x - 15 = 0
4x(x + 3) - 5(x+3) = 0
(4x - 5) (x +3) = 0
x = 5/4 or x = -3

Answer 2

To complete the square, first divide by 4 so you have:
x^2 + 7/4x = 15/4
Then add (b/2)^2 where b is the coefficient on the linear term, 7/4 in this case.
x^2 + 7/4x + 49/64 = 15/4 + 49/64
Now factor the quadratic on the left:
( x + 7/8 )^2 = 289/64

Answer 3

ax²+bx+c=0
1st)Move the c to the right hand side(remain unmoved if c is
on the right hand side initially like the question).
2nd)Factorise out the a to make it to 1x² and so do the value
of b by dividing b by a.
3rd)Divide c by a factorised.
4th)Write the terms left on the left as (x+k/2)²-k² where k is a
constant.
5th)Carry out normal computation.Do REMEMBER to write ±
as you surd any value in an quadratic equation as it has
got 2 roots.
4x²+7x=15
4(x²+7/4x)=15
x²+7/4x=15/4
(x+7/8)²-49/64=15/4
(x+7/8)²=15/4+49/64
(x+7/8)²=289/64
x+7/8=±√(289/64)
x+7/8=±(17/8)
x+7/8=17/8 , x+7/8=-17/8
x=5/4 x=-3

Answer 4

4x^2 + 7x = 15

divide by 4

x^2 + (7/4)x = 15/4

x^2 + 2(7/8) x + (7/8)^2 = 15/4 + (7/8)^2

=>(x +7/8)^2 = 15/4 + 49/64

=>(x+7/8)^2 = 289/64

=>(x+7/8)^2 = (17/8)^2

x + 7/8 = ± 17/8

x = -7/8 ± 17/8

x = 10/8 or -24/8

x = 5/4 or -3

Answer 5

4 ( x ² + 7/4 x ) = 15
x ² + (7/4) x = 15/4
x ² + (7/4) x + 49/64 = 15/4 + 49/64
(x + 7/8) ² = (240 + 49) / 64
(x + 7/8) ² = 289 / 64
x + 7/8 = ± 17 / 8
x = - 7/8 ± 17/8
x = (-1/8) (7 ± 17)
x = (-1/8)(24) , x = 10/8
x = - 3 , x = 5 / 4

Answer 6

Take a look at this

http://www.purplemath.com/modules/sqrquad.htm