How do you solve $2 x^{2} - 7 x - 3 = 0$ $2x^2 - 7x - 3 = 0$ ?

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How do you solve $2 x^{2} - 7 x - 3 = 0$ $2x^2 - 7x - 3 = 0$ ?

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Answer 1 · 2015-09-09T11:37:09

$x = \frac{7 \pm \sqrt{73}}{4}$ $x=(7+-sqrt(73))/4$

Explanation:

An examination of the factors of $2$ $2$ and $- 3$ $-3$ reveals that there are no integer solutions.

Therefore, we will use the quadratic root formula
$XXX x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $color(white)("XXX")x= (-b+-sqrt(b^2-4ac))/(2a)$
$XXXXXXXXXXX$ $color(white)("XXXXXXXXXXX")$ for a quadratic in the general form $a x^{2} + b x + c = 0$ $ax^2+bx+c=0$

In this case
$XXX x = \frac{7 \pm \sqrt{7^{2} - 4 (2) (- 3)}}{2 (2)}$ $color(white)("XXX")x = (7+-sqrt(7^2-4(2)(-3)))/(2(2))$

$XXXX = \frac{7 \pm \sqrt{49 + 24}}{4}$ $color(white)("XXXX")=(7+-sqrt(49+24))/4$

$XXXX = \frac{7 \pm \sqrt{73}}{4}$ $color(white)("XXXX")=(7+-sqrt(73))/4$

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How do you solve $2 x^{2} - 7 x - 3 = 0$ $2x^2 - 7x - 3 = 0$ ?

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