# How do you solve 4x^2+23x-10=10x + x^2 by using the quadratic formula?

Mar 7, 2017

$x = - 5$ or $x = \frac{2}{3}$

#### Explanation:

$4 {x}^{2} + 23 x - 10 = 10 x + {x}^{2}$

$3 {x}^{2} + 13 x - 10 = 0$

$x = \frac{- 13 \pm \sqrt{{13}^{2} - 4 \times 3 \times \left(- 10\right)}}{2 \times 3}$

$= \frac{- 13 \pm \sqrt{169 + 120}}{2 \times 3}$

$= \frac{- 13 \pm \sqrt{289}}{6}$

$= \frac{- 13 \pm 17}{6}$

$= - 5$ or $\frac{2}{3}$

But also consider:

$3 {x}^{2} + 13 x - 10 = 0$

$\left(x + 5\right) \left(3 x - 2\right) = 0 \to x = - 5$ or $\frac{2}{3}$

Therefore we could have arrived at the result without the quadratic formula in this case.