# solve and explain, thank you: 3x^2= x+4 ; 25= 9x^2 - 30x

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(1) Solve 3x^2=x+4
3x^2-x-4=0 Bring all terms to one side of the equation -- put in standard form.
(3x-4)(x+1)=0 Factor. There are many techniques -- here you could use a guess and check method. The leading coefficient is 3, which is prime, so the first terms in the parantheses must be 3x and 1x. The second numbers in the parantheses must have a product of -4. The only possibilities to try are +/-2, -1 and 4, 1 and negative
4.
Using the zero product property (if a,b are real numbers and ab=0, then either a=0,b=0, or both are zero):
3x-4=0 implies x=4/3
x+1=0 implies x=-1
So the two solutions are 4/3 or -1
(2) 25=9x^2-30x
We follow the steps above:
9x^2-30x-25=0
Using the quadratic formula we get x=5/3+5/3sqrt(2) or 5/3 - 5/3sqrt(2)
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If there is a typo in the question and you meant:
-25=9x^2-30x then
9x^2-30x+25=0 is a special product pattern: a^2-2ab+b^2=(a-b)^2
Here a=3x,b=5 so:
9x^2-30x+25=0
(3x-5)(3x-5)=0
the solution is x=5/3

`3x^2= x+4` ;

`25= 9x^2 - 30x `

`3x^2= x+4 `

`3x^2-x-4 `

`3xx-4=-12 ` factors that add to -12 = -4 and 3

`3x^2+3x-4x-4 `

factors

`(3x^2+3x)(-4x-4) `

`3x(x+1)-4(x+1) `

`(3x-4)(x+1) `

`3x-4=0 `

`x=4/3` or `1 1/3`

x+1=0

x=1

`-25= 9x^2 - 30x `

`9x^2-30x+25 `

find the square root of a and c

`(3x+5)(3x+5) `

`x=-5/3`