# How do you solve -4x^2+4x=9?

Jan 21, 2015

For a general quadratic $a {x}^{2} + b x + c$, you know that the solutions are given by ${x}_{1 , 2} = \setminus \frac{- b \setminus \pm \setminus \sqrt{{b}^{2} - 4 a c}}{2 a}$.

First of all, we rewrite the equation as $- 4 {x}^{2} + 4 x - 9 = 0$, which (if you prefer) can be also written as $4 {x}^{2} - 4 x + 9 = 0$

In this final case, $a = 4$, $b = - 4$, $c = 9$. Substituing this values into the solving formula, we get

${x}_{1 , 2} = \setminus \frac{4 \setminus \pm \setminus \sqrt{16 - 4 \setminus \cdot 4 \setminus \cdot 9}}{2 \setminus \cdot 4}$

The discriminant ${b}^{2} - 4 a c$ is negative, and thus there are no real solutions.

If you're interested in complex solutions, you get
${x}_{1 , 2} = \setminus \frac{4 \setminus \pm i \setminus \sqrt{128}}{8} = \setminus \frac{4 \setminus \pm 8 i \setminus \sqrt{2}}{8}$
which yelds
${x}_{1} = \setminus \frac{1}{2} + i \setminus \sqrt{2}$,
${x}_{2} = \setminus \frac{1}{2} - i \setminus \sqrt{2}$.