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

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Answer 1 · 2018-05-12T03:06:13

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

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

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

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

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

$x = \frac{2 \pm \sqrt{4 + (4 \cdot 8 \cdot 5)}}{2 \cdot 8}$ $x = (2 +- sqrt(4 + (4 * 8 * 5))) / (2 * 8)$

$x = \frac{2 \pm \sqrt{164}}{16}$ $x = (2 +- sqrt(164)) / 16$

$x = \frac{1 \pm \sqrt{41}}{8}$ $color(blue)(x = (1 +- sqrt(41)) / 8$

How do you solve 8x2−2x−5=08x^2-2x-5=0 using the quadratic formula?