# Solve for $x$, $-\frac{1}{2}x^2 + 2x + 5 = 0$

I'm having trouble solving this equation for $x$:

$$-\frac{1}{2}x^2 + 2x + 5 = 0$$

What's the steps to take to solve it?

Thanks.

• The equation in the question.. – Cypras Feb 7 '13 at 23:02
• Not sure how to format properly? It should be -1/2 – Cypras Feb 7 '13 at 23:04
• @Cypras: the equation is not written properly and it is not clear what you meant as the LaTex is a bit hosed. Regards – Amzoti Feb 7 '13 at 23:04
• Are you familiar with the quadratic formula? – Gerry Myerson Feb 7 '13 at 23:05
• Multiply by -2 and use the ABC formula – Slugger Feb 7 '13 at 23:05

$$x^2-4x-10=0$$ $$x^2-2\cdot x\cdot2+2^2-2^2-10=0$$ $$x^2-2\cdot x\cdot2+2^2=4+10$$ $$(x-2)^2=14$$ $$x-2=\pm \sqrt{14}$$ $$x=2\pm \sqrt{14}$$

• +1 simple but leading approach. – mrs Feb 8 '13 at 17:39
• @Thank you very much Babak. – Adi Dani Feb 8 '13 at 18:30

If you multiply by $-2$ you get $x^2-4x-10=0$, then you can use the quadratic formula to get $x=\frac 12 (4 \pm \sqrt {16+40})=2 \pm \sqrt {14}$

• How did you get 14? – Cypras Feb 7 '13 at 23:11
• sqrt(16 + 40) = sqrt(56) = sqrt(2 * 2 * 14) = 2 * sqrt(14) – oks Feb 7 '13 at 23:14
• @oks: I am not sure OP will see your comment. You can preface it with <at>Cypras as Rustyn Yazdanpour did to make sure s/he does. You can only notify one user with the at sign per comment. – Ross Millikan Feb 7 '13 at 23:18
• @Rustyn That makes no sense to me. Square root of 56 equals 7.4something? How do you change it like that? – Cypras Feb 7 '13 at 23:20
• @Cypras: Rustyn Yazdanpour is doing arithmetic under the square root sign, then using the fact that $\sqrt{ab}=\sqrt a \cdot \sqrt b$. The $40$ in the next to last should be $14$. – Ross Millikan Feb 7 '13 at 23:26

If you multiply $$-\frac{1}{2}x^2 + 2x + 5 = 0\tag{1}$$

by $-2$ you get $$x^2-4x-10=0$$ Using the quadratic formula: $$ax^2 + bx + c = 0 \iff x = \frac{1}{2a}\left(-b \pm \sqrt{b^2 - 4ac}\right),$$ where in this case, $a = 1,\;b= -4,\; c = -10$ we have $$x=\frac 12 (4 \pm \sqrt {16+40})=2 \pm \sqrt {14}\tag{2}$$

Walking through the simplification of $(2)$

$$x = \frac12 (4 \pm \sqrt{16 + 40}) \;=\; \frac12(4 \pm \sqrt{56})\; = \;\frac12 (4 \pm \sqrt{4\cdot 14}) \;=\; \frac12 (4 \pm \sqrt{4}\sqrt{14})\;$$ $$= \;\frac12 (4 \pm 2\sqrt{14})\;=\; \frac 12 \cdot 4 \pm \frac12 \cdot 2\sqrt{14}\;\; = \;\;2 \pm \sqrt{14}$$

• We, here, denote the discriminate by $\Delta=b^2-4ac$. + – mrs Feb 9 '13 at 5:34