+0

# Help 20 ​

0
268
6

Help 20 Nov 23, 2018

#1
+2

We can solve this using quadratic formula, $$x = {-b \pm \sqrt{b^2-4ac} \over 2a}$$ . (Glad that web2.0calc gives you this). Plugging the numbers into the quadratic formula, we have one solution as $$x=\frac{-3+\sqrt{3^2-4\cdot \:1\left(-5\right)}}{2\cdot \:1}, \quad \frac{-3+\sqrt{29}}{2}$$ and the other solution as $$x=\frac{-3-\sqrt{3^2-4\cdot \:1\left(-5\right)}}{2\cdot \:1}:\quad \frac{-3-\sqrt{29}}{2}.$$ Thus, the final solutions as $$x=\frac{-3+\sqrt{29}}{2},\:x=\frac{-3-\sqrt{29}}{2}.$$

.
Nov 23, 2018

#1
+2

We can solve this using quadratic formula, $$x = {-b \pm \sqrt{b^2-4ac} \over 2a}$$ . (Glad that web2.0calc gives you this). Plugging the numbers into the quadratic formula, we have one solution as $$x=\frac{-3+\sqrt{3^2-4\cdot \:1\left(-5\right)}}{2\cdot \:1}, \quad \frac{-3+\sqrt{29}}{2}$$ and the other solution as $$x=\frac{-3-\sqrt{3^2-4\cdot \:1\left(-5\right)}}{2\cdot \:1}:\quad \frac{-3-\sqrt{29}}{2}.$$ Thus, the final solutions as $$x=\frac{-3+\sqrt{29}}{2},\:x=\frac{-3-\sqrt{29}}{2}.$$

tertre Nov 23, 2018
#2
+2

Excellent, tertre...your LaTex presentations are good...   CPhill  Nov 23, 2018
edited by CPhill  Nov 23, 2018
#3
+1

Thank you, CPhill! I have been working on them.

tertre  Nov 23, 2018
#4
+2

umm i'm kinda confused still :(

NotSoSmart  Nov 24, 2018
#5
+2

tertre  Nov 24, 2018
#6
+1

1x^2 + 3x - 5   = 0

We have the form   ax^2 + bx + c   = 0

Where a = 1     b = 3      and    c   = -5

[  - b ± √ [  b^2  - (4 * a * c) ]   ]     /    [ 2 * a ]

Plugging in for a, b, c......we have

[ - 3 ± √ [ 3^2 - (4 * 1 * -5) ] ]  /   [ 2 * 1 ]       simplify

[ -3  ± √ [ 9 - ( -20) ] } /  2

[ - 3  ± √ 29   ]    /   2

And the two solutions as are tertre found

x = [ -3  + √ 29   ]    /   2        and       x =    [- 3  - √ 29   ]    /   2   Nov 26, 2018