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# How would i go about solving?

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How would i go about solving?

7+- the square root of 9 divided by 2

I got it using the quadratic formula from x^2+4x-7=0

Guest May 8, 2017
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Your question is worded really well .. I can see you have tried it properly for yourself.  I like that :)

7+- the square root of 9 divided by 2

I got it using the quadratic formula from x^2+4x-7=0

$$x = {-b \pm \sqrt{b^2-4ac} \over 2a}\\ x = {-4 \pm \sqrt{4^2-4*1*-7} \over 2*1}\\ x = {-4 \pm \sqrt{16--28} \over 2}\\ x = {-4 \pm \sqrt{16+28} \over 2}\\ x = {-4 \pm \sqrt{44} \over 2}\\ x = {-4 \pm \sqrt{4*11} \over 2}\\ x = {-4 \pm2 \sqrt{11} \over 2}\\ x = {2(-2 \pm \sqrt{11} )\over 2}\\ x=-2 \pm \sqrt{11}$$

So if you did not mistype your original equation then you did something wrong with the quadratic formula.

Now I will look at you ACTUAL question ... which is simpler :)

7+- the square root of 9 divided by 2

$$x = {7 \pm \sqrt{9} \over 2}\\ x = {7 \pm 3 \over 2}\\ x = {7 + 3 \over 2} \qquad or \qquad x = {7 - 3 \over 2}\\ x = {10 \over 2} \qquad \;\; or \qquad x = {4 \over 2}\\ x = 5 \qquad \quad or\quad \quad x = 2\\$$

This would have been the solution to      $$x^2-7x+10=0$$

Melody  May 8, 2017

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