How do you solve $4 x^{2} - 15 x = 3$ $4 x ^2 − 15 x = 3$ ?

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How do you solve $4 x^{2} - 15 x = 3$ $4 x ^2 − 15 x = 3$ ?

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Answer 1 · 2018-05-09T18:56:40

$x = \frac{15 \pm \sqrt{273}}{8}$ $x=(15+-sqrt273)/8$

Explanation:

$4 x^{2} - 15 x = 3$ $4x^2-15x=3$

$4 x^{2} - 15 x - 3 = 0$ $4x^2-15x-3=0$

Use the quadratic formula

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

with $a = 4$ $a=4$ , $b = - 15$ $b=-15$ , $c = - 3$ $c=-3$

$x = \frac{- (- 15) \pm \sqrt{{(- 15)}^{2} - 4 (4) (- 3)}}{2 (4)} = \frac{15 \pm \sqrt{225 + 48}}{8} = \frac{15 \pm \sqrt{273}}{8}$ $x=(-(-15)+-sqrt((-15)^2-4(4)(-3)))/(2(4))=(15+-sqrt(225+48))/8=(15+-sqrt273)/8$

Answer 2 · 2018-05-09T19:05:45

$4 x^{2} - 15 x = 3$ $4x^2-15x=3$
$4 x^{2} - 15 x - 3 = 0$ $4x^2-15x-3=0$
$D= β^2-4αγ$
$D= 15^2-4*4*-3$
$D= 225+48$
$D= 273$

$x1,2=(-β+- sqrt273)/(2α)$
$x1,2=(15+- sqrt273)/8$
$x1=(15+ sqrt273)/8$
$x2=(15- sqrt273)/8$

Explanation:

You use the Discriminant. That's how it works: https://www.khanacademy.org/math/algebra/quadratics/solving-quadratics-using-the-quadratic-formula/a/discriminant-review

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How do you solve $4 x^{2} - 15 x = 3$ $4 x ^2 − 15 x = 3$ ?

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