How do I solve this by factorizing?

Question

2x^2+3=5x+1 also 3x^2 + 5x

How do I solve this by factorizing?

EDIT :also 3x^2 + 5x = 0

Answer 1

1. 2x^+3=5x+1
=>2x^2+3-5x-1=0
=>2x^2-5x+2=0
=>2x^2-4x-x+2=0
=>2x(x-2)-1(x-2)=0
=>(x-2)(2x-1)=0
Therefore ,either x-2=0 or 2x-1=0
Therefore,x=2 or 1/2 ans
2. 3x^2+5x=0
=>x(3x+5)=0
=>Either x=0 or 3x+5=0
Therefore x=0 or -5/3

Answer 2

Easy: First:
3x^2 + 5x
=x(3*x + 5)=0
so either x=0 or 3*x + 5=0
in this case, 3*x+5=0, 3x=-5, so x=-5/3.
So the possible solitons are x=-5/3 and x=0

Second:
2x^2+3=5x+1
2x^2-5*x+2=0
(x-2)*(2*x-1)=0
so either x-2=0 or 2*x-1=0
then x=2 or x=1/2

Answer 3

2x^2+3=5x+1
move it to one side... 2x^2 -5x +2 = 0
well, it doesn't factor normally, but you can do the squaring method (make the first side a binomial squared and set it equal to what's left)
first, divide by 2: x^2 - 5/2x +1 = 0
add a number to each side: x^2 - 2.5x + (5/4)^2 = (5/4)^2 - 1
(x - 5/4)^2 = 25/16 - 16/16 = 9/16
square root it: x - 1.25 = square root of 9/16 (which is plus or minus 3/4)
x = 1.25 + or - 0.75
x = 2 and 0.5

the next one's simple enough: 3x^2 + 5x = 0
factor out x: (x)(3x + 5) = 0
set each ()-thing to zero...
x = 0
3x + 5 = 0 --> 3x = -5 --> x = -5/3
x = 0 and -5/3

Answer 4

2x^2 + 3 = 5x + 1
2x^2 - 5x + 2 = 0

x = (-b ± sqrt(b^2 - 4ac))/(2a)

x = (-(-5) ± sqrt(25 - 4(2)(2)))/(2(2))
x = (5 ± sqrt(25 - 16))/4
x = (5 ± sqrt(9))/4
x = (5 ± 3)/4
x = (8/4) or (2/4)
x = 2 or (1/2)

ANS : (x - 2)(2x - 1)

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3x^2 + 5x = 0
x(3x + 5)

Assuming you didn't leave anything out.