How do you solve $8x^2-2x-5=0$ using the quadratic formula?

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Answer 1 · 2018-05-12T03:06:13

$color(blue)("Two roots are " x = (1 + sqrt(41)) / 8, (1 - sqrt(41)) / 8$

$"Quadratic Formula : "x = (-b +- sqrt(b^2 - 4ac)) / (2a)$

$"Given " 8x^2 - 2x - 5 = 0$

$a = 8, b = -2, c = -5$

Substituting values of a, b, c in the quadratic formula,

$x = (2 +- sqrt(4 + (4 * 8 * 5))) / (2 * 8)$

$x = (2 +- sqrt(164)) / 16$

$color(blue)(x = (1 +- sqrt(41)) / 8$

How do you solve 8x^2-2x-5=08x^2-2x-5=0 using the quadratic formula?