## Diana.xL one year ago http://prntscr.com/8m4yb0

1. Diana.xL

@ganeshie8

2. Diana.xL

@phi

3. phi

do you see the 5x term? take the 5 from that term, divide it by 2 $\frac{5}{2}$ then square it $\frac{25}{4}$ add that to both sides of the equation

4. phi

what do you get?

5. Diana.xL

6. phi

what do you get after adding 25/4 to both sides of the equation?

7. phi

you should get $x^2 +5x + \frac{25}{4}= -1 + \frac{25}{4}$

8. phi

the left side is a perfect square (that is why you add 25/4) it is $\left(x+ \frac{5}{2}\right)^2 = -1 + \frac{25}{4}$ I would write the -1 as -4/4 $\left(x+ \frac{5}{2}\right)^2 = \frac{-4}{4} + \frac{25}{4}$

9. phi

on the right side you have two fractions with the same denominator, so you can add the tops

10. Diana.xL

21/4

11. phi

yes $\left(x+ \frac{5}{2}\right)^2 = \frac{21}{4}$ if you take the square root (of both sides to keep it equal) we get rid of the square on the left side $\left(x+ \frac{5}{2}\right)= \pm \sqrt\frac{21}{4} \\ x+ \frac{5}{2}= \pm\frac{\sqrt{21}}{2}$ now add -5/2 to both sides

12. phi

you get $x= \frac{-5 \pm \sqrt{21}}{2}$