# How do you solve -3x^2=2(3x-5) using the quadratic formula?

Jan 22, 2017

Rearrange to form: $3 {x}^{2} + 6 x - 10 = 0$
Substitute in quadratic formula: $x = \frac{- \left(6\right) \pm \sqrt{{\left(6\right)}^{2} - 4 \left(3\right) \left(- 10\right)}}{2 \left(3\right)}$
Simplify to get: $x = \frac{- 3 + \sqrt{39}}{3} \approx 1.08167$ or $x = \frac{- 3 - \sqrt{39}}{3} \approx - 3.08167$

#### Explanation:

First of all you need to rearrange the equation into the form $a {x}^{2} + b x + c = 0$

So first expand the brackets on the right hand side: $- 3 {x}^{2} = 6 x - 10$

Then you can either take the $- 3 {x}^{2}$ over or the $6 x - 10$ over.

Let's just take the $- 3 {x}^{2}$ over to give: $3 {x}^{2} + 6 x - 10 = 0$

Now we can see what $a$ $b$ and $c$ are:
$a = 3$
$b = 6$
$c = - 10$

Then we substitute these into the quadratic formula:
$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$
Quadratics usually have 2 answers which is why there is a $\pm$ as for one answer you use $+$ and the other $-$.

So now substituting in: $x = \frac{- \left(6\right) \pm \sqrt{{\left(6\right)}^{2} - 4 \left(3\right) \left(- 10\right)}}{2 \left(3\right)}$

This simplifies to: $x = \frac{- 6 \pm 2 \sqrt{39}}{6}$

And dividing all terms by 2 to: $x = \frac{- 3 \pm \sqrt{39}}{3}$ which is the most simple form.

So $x = \frac{- 3 + \sqrt{39}}{3} \approx 1.08167$ or $x = \frac{- 3 - \sqrt{39}}{3} \approx - 3.08167$