# How do you solve 30 + x − x^2 = 0?

Aug 1, 2015

$x = - 5 , 6$

#### Explanation:

Invert (multiply by -1, has the same solutions) and complete the square:
${x}^{2} - x - 30 = {\left(x - \frac{1}{2}\right)}^{2} - \frac{121}{4} = 0$

Solve for $x$:
${\left(x - \frac{1}{2}\right)}^{2} = \frac{121}{4}$

=>

$x - \frac{1}{2} = \pm \frac{11}{2}$

=>

$x = \frac{1 \pm 11}{2}$

Aug 1, 2015

solve $y = - {x}^{2} + x + 30 = 0$

Ans: -5 and 6

#### Explanation:

I use the new Transforming Method (Google, Yahoo, Bing Search)
Find 2 numbers knowing sum (1) and product (-30). Roots have opposite signs since a and c have opposite signs.
Factor pairs of (-30) --> (-2, 15)(-4, 5)(-5, 6). This sum is 1 = b.
Since a < 0. then the 2 real roots are: -5 and 6.

Aug 1, 2015

You could use the quadratic formula.

#### Explanation:

$\textcolor{b l u e}{a {x}^{2} + b x + c = 0}$

for which the quadratic formula takes the form

color(blue)(x_(1,2) = (-b +- sqrt(b^2 - 4ac))/(2a)

You will start from

$- {x}^{2} + x + 30 = 0$

which can be rewritten as

$- \left({x}^{2} - x - 30\right) = 0$

In this case, $a = 11$, $b = - 1$, and $c = - 30$.

The two solutions to this quadratic equation will thus be

${x}_{1 , 2} = \frac{- \left(- 1\right) \pm \sqrt{{\left(- 1\right)}^{2} - 4 \cdot \left(1\right) \cdot \left(- 30\right)}}{2 \cdot \left(1\right)}$

${x}_{1 , 2} = \frac{1 \pm \sqrt{121}}{- 2} = \frac{1 \pm 11}{2}$

${x}_{1} = \frac{1 + 11}{2} = \textcolor{g r e e n}{6}$

${x}_{2} = \frac{1 - 11}{2} = \textcolor{g r e e n}{- 5}$