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Its all good i forgive you!

Great job Heureka!  You're smart!

(2x-10)(x+5)=x^2+10x-75

okay:  x^2-10x=-25

now: $$x^2-10x+25=0$$

use the abc - formula:

if $$ax^2+bx+c = 0$$ than $$x_{1,2} = \frac{1}{2a}*(b-\sqrt{b^2-4*a*c})$$

$$1x^2 -10x+25=0 \\
\underbrace{1}_{a=1}x^2 \underbrace{-10}_{ b=-10}x+\underbrace{25}_{c=25}=0 \\\\
x_{1,2} = \frac{1}{2*1}*(10-\sqrt{100-4*1*25})\\\\
x_{1,2} = \frac{1}{2}*(10-\sqrt{100-100})\\\\
x = \frac{1}{2}*(10-0)\\\\
x = \frac{1}{2}*10\\\\
x = \frac{10}{2}\\\\
x = 5$$

Thank you happy:)

I forgive you....

All you people give her another chance, please?

Why thank you!

wow you rock

Exclamation point expresses excitment at the end of a sentence, it is a piece of punctuation

19 1/2=39/2

3 3/4= 15/4

(39/2)/(15/4)  Let's flip 15/4 to 4/15 then multiply

(39/2)*(4/15)  that's the new sentence

39*4=156

2*15=30

156/30=26/5

26/5=5 1/5

Just to check my work:

$${\frac{\left({\frac{{\mathtt{39}}}{{\mathtt{2}}}}\right)}{\left({\frac{{\mathtt{15}}}{{\mathtt{4}}}}\right)}} = {\frac{{\mathtt{26}}}{{\mathtt{5}}}} = {\mathtt{5.2}}$$

So I am correct and there you go!

262464195387

$${\frac{\left({\frac{{\mathtt{5}}}{{\mathtt{6}}}}\right)}{\left({\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)}} = {\frac{{\mathtt{5}}}{{\mathtt{2}}}} = {\mathtt{2.5}}$$

yep thats it

262464195387

You'll have to be more specific.....we aren't magical

your welcome

Thanks!!!!!!!!

 I agree with happy7 he is awesome at math!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

...262464195387.

Well 12*12=144

So 3/4 of 12 would be 9

Is that what you mean?

And the answer is 90%

Heureka did this partly by visual inspection - it made sense to do it that way.

I always look for the L O N G E S T solution. :)     LOL

$$\\\sqrt{\sqrt{x-5} + x} = 5\\\\
\left[\sqrt{\sqrt{x-5} + x}\right]^2 = 5^2\\\\
\sqrt{x-5} + x = 25\\\\
\sqrt{x-5} = 25-x\\\\
(\sqrt{x-5})^2 = (25-x)^2\\\\
x-5= 625+x^2-50x \\\\
x^2-51x+630=0\\\\$$

$$\\\triangle=51^2-4*1*630=81\\\\
x=\frac{51\pm 9}{2}\\\\
x=\frac{51+ 9}{2}\qquad or\qquad x=\frac{51- 9}{2}\\\\
x=\frac{60}{2}\qquad or\qquad x=\frac{42}{2}\\\\
x=30 \qquad or\qquad x=21\\\\$$

Test

x=30      

$$\\\sqrt{\sqrt{30-5}+30}\\
=\sqrt{5+30}\\
=\sqrt{35}\\
\ne5$$        x=30 is not a solution

Test x=21

$$\\\sqrt{\sqrt{21-5}+21}\\
=\sqrt{4+21}\\
=\sqrt{25}\\
=5\qquad excellent$$

So  the answer is x=21

sqrt(sqrt(x-5) + x) = 5        x = 21

$$\sqrt{ \sqrt{x-5 } +x } = 5 \quad | \quad x = 21\\\\
\sqrt{ \sqrt{21-5 } +21 } = 5 \\\\
\sqrt{ \sqrt{16 } +21 } = 5 \\\\
\sqrt{ 4 +21 } = 5 \\\\
\sqrt{ 25 } = 5 \\\\
5 \stackrel{!}{=} 5 \\$$

Did you see that I mentioned your great answer in my wrap tonight  :)

The address of the wrap is in the "lantern thread".  You can jump over to it from there :)

Thank you Melody! 

Actually I'm just about to be a Primary 5 student... Anyway thanks again!

Regards,