How do you solve $3 x^{2} - 8 = 0$ $3x^2-8=0$ using the quadratic formula?

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Answer 1 · 2017-07-17T21:34:08

See a solution process below:

We can rewrite this equation as:

$3 x^{2} + 0 x - 8 = 0$ $3x^2 + 0x - 8 = 0$

For $a x^{2} + b x + c = 0$ $ax^2 + bx + c = 0$ , the values of $x$ $x$ which are the solutions to the equation are given by:

$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $x = (-b +- sqrt(b^2 - 4ac))/(2a)$

Substituting $3$ $3$ for $a$ $a$ ; $0$ $0$ for $b$ $b$ and $- 8$ $-8$ for $c$ $c$ gives:

$x = \frac{- 0 \pm \sqrt{0^{2} - (4 \cdot 3 \cdot - 8)}}{2 \cdot 3}$ $x = (-0 +- sqrt(0^2 - (4 * 3 * -8)))/(2 * 3)$

$x = \pm \frac{\sqrt{0 - (- 96)}}{6}$ $x = +- sqrt(0 - (-96))/(6)$

$x = \pm \frac{\sqrt{+ 96}}{6}$ $x = +- sqrt(+96)/(6)$

$x = \pm \frac{\sqrt{16 \cdot 6}}{6}$ $x = +- sqrt(16 * 6)/(6)$

$x = \pm \frac{\sqrt{16} \cdot \sqrt{6}}{6}$ $x = +- (sqrt(16) * sqrt(6))/(6)$

$x = \pm \frac{4 \cdot \sqrt{6}}{6}$ $x = +- (4 * sqrt(6))/6$

$x = \pm \frac{2 \cdot 2 \cdot \sqrt{6}}{2 \cdot 3}$ $x = +- (2 * 2 * sqrt(6))/(2 * 3)$

$x = +- (color(red)(cancel(color(black)(2))) * 2 * sqrt(6))/(color(red)(cancel(color(black)(2))) * 3)$

$x = +- (2sqrt(6))/3$

How do you solve 3x^2-8=03x2−8=03x^2-8=0 using the quadratic formula?