# How do you solve the quadratic 1-6x^2=-15 using any method?

Sep 3, 2016

$x = \frac{2 \sqrt{6}}{3}$ or $x = \frac{- 2 \sqrt{6}}{3}$

#### Explanation:

$1 - 6 {x}^{2} = - 15$

$\Leftrightarrow 6 {x}^{2} - 15 - 1 = 0$ or

$6 {x}^{2} - 16 = 0$

According to quadratic formula solution of equation ax^2+bx×c=0 is given by

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

Hence, solution of $6 {x}^{2} - 16 = 0$ is

x=(-0+-sqrt(0^2-4×6×(-16)))/(2×6)

= +-sqrt(4×6×16)/12

= $\pm 8 \frac{\sqrt{6}}{12}$

= $\pm \frac{2 \sqrt{6}}{3}$

Hence solution of equation is $x = \frac{2 \sqrt{6}}{3}$ or $x = \frac{- 2 \sqrt{6}}{3}$