A little confused on this problem.. 4x^2=12x+40

I have to use standard form to solve..

4x^2-12x-40=0

your first dtep is correct :) do you have the standard formula or should i mention it ?

*step

add 40 to the 0 that leaves 4x^2 -12x = 40

4(x^2-3X+10)=0 (x-5)(x+2)=0 x=5 or x=-2

whats actually the confusion about ?

4(x^2-3X+10)=0 (x-5)(x+2)=0 x=5 or x=-2

how did you get the (x-5)(x+2)

@beesknees01 , if you need to use standard formula, then you should not be concerned about how (x-5)(x+2), you'll get the roots directly using the standard formula

@hartnn yea he is right use standard formula

here it goes, Compare your quadratic equation with \(ax^2+bx+c=0\) find \[a=...?\\b=...?\\c=...?\] then the two roots of x are: \(\huge{x_{1,2}=\frac{-b \pm \sqrt{b^2-4ac}}{2a}}\)

It's just in the lesson, the steps are 1. Write in standard 2. Factor quadratic expression 3. set each factor to 0

so, we don't need the standard formula here, we can do it by factorization. for ax^2+bx+c=0, we search for 2 numbers with sum = b and product =ac so, for your question, first factor out 4, 4(x^2-3X+10)=0 (x^2-3X+10)=0 what are 2 numbers with sum =-3 and product =10 ??

i mean product = -10

x=3 and x=10?

oops, nevermind. That was wrong.

can you think of 2 numbers with sum =-3 and product =-10 ??

Im just putting it in there make sure you know the standard formula Ax^2 + Bx + C = 0 so you can apply it to the quadratic formula.

hint : one of the numbers will be negative.

-5 and 2

those are correct! so, we split -3x as -5x+2x we have x^2-5x+2x-10 =0 can you factor out, x from first 2 terms and +2 from other 2 terms ?

x^2-3x-10=0

from this : x^2-3x-10=0 we got this :x^2-5x+2x-10=0 now you can factor out 'x' from x^2-5x, what you get ?

example : if there was a^2+2ab when i factor out 'a' i get a (a+2b)

(x+2)(x-5)=0?

you got how we get (x+2)(x-5)=0 ?? because thats correct...

i found factors that have a sum of -3 and when multiplied equals -10

ok, last step will be to equate them to 0 x+2 =0 , x-5 =0 x = -2, x=5 those are the values of x you require.

any more doubts ?

Nope, thank you very much!!

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