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

Question

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.

Answer 1

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

Answer 2

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

Answer 3

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 $(2)$

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