# How do you solve 9x^2-5x=7?

Aug 8, 2015

Subtract $7$ from both sides to get: $9 {x}^{2} - 5 x - 7 = 0$, then solve using the quadratic formula to get:

$x = \frac{5 \pm \sqrt{277}}{18}$

#### Explanation:

$9 {x}^{2} - 5 x - 7$ is of the form $a {x}^{2} + b x + c$, with $a = 9$, $b = - 5$ and $c = - 7$.

This has discriminant $\Delta$ given by the formula:

$\Delta = {b}^{2} - 4 a c = {\left(- 5\right)}^{2} - \left(4 \times 9 \times - 7\right) = 25 + 252$

$= 277$

Since $\Delta > 0$, the quadratic equation has two distinct Real roots. Since $\Delta$ is not a perfect square ($277$ is prime), those roots are irrational.

The roots are given by the quadratic formula:

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

$= \frac{- b \pm \sqrt{\Delta}}{2 a}$

$= \frac{5 \pm \sqrt{277}}{18}$

Notice the discriminant $\Delta$ is the expression under the square root.

So if $\Delta < 0$ the square root is not Real and the quadratic has no Real roots - It has a conjugate pair of distinct complex roots.

If $\Delta = 0$ then there is one repeated Real root.

If $\Delta > 0$ (as in our example), there are two distinct Real roots. If in addition $\Delta$ is a perfect square, then those roots are rational.