How do you solve $5 x^{2} - 3 x - 3 = 0$ $5x^2-3x-3=0$ using the quadratic formula?

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Answer 1 · 2016-05-22T08:30:02

$x = \frac{3 + \sqrt{69}}{10}$ $x=(3+ sqrt(69))/10$

$x = \frac{3 - \sqrt{69}}{10}$ $x=(3- sqrt(69))/10$

The quadratic formula is $\frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $(-b± sqrt(b^2-4ac))/(2a)$

In the equation $5 x^{2} - 3 x - 3 = 0$ $5x^2-3x-3=0$

$a = 5$ $a=5$
$b = - 3$ $b=-3$
$c = - 3$ $c=-3$

Therefore,

$\frac{3 \pm \sqrt{- 3^{2} - 4 \times 5 \times - 3}}{2 \times 5}$ $(3± sqrt(-3^2-4xx5xx-3))/(2xx5)$

= $\frac{3 \pm \sqrt{9 - - 60}}{10}$ $(3± sqrt(9--60))/(10)$

$x = \frac{3 \pm \sqrt{69}}{10}$ $x=(3± sqrt(69))/10$

How do you solve 5x2−3x−3=05x^2-3x-3=0 using the quadratic formula?