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

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Answer 1 · 2016-05-22T08:30:02

$x=(3+ sqrt(69))/10$

$x=(3- sqrt(69))/10$

The quadratic formula is $(-b± sqrt(b^2-4ac))/(2a)$

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

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

Therefore,

$(3± sqrt(-3^2-4xx5xx-3))/(2xx5)$

= $(3± sqrt(9--60))/(10)$

$x=(3± sqrt(69))/10$

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