How do you solve $20 x^{2} = 13 - x$ $20x^2 = 13 - x$ ?

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How do you solve $20 x^{2} = 13 - x$ $20x^2 = 13 - x$ ?

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Answer 1 · 2018-04-07T20:54:54

$x \approx 0.78, 0.83$ $x~~0.78,0.83$

Explanation:

Rearranging the equation
$\Rightarrow 20 x^{2} + x - 13 = 0 \Rightarrow (1)$ $=>20x^2+x-13=0 => (1)$
$a x^{2} + b x + c = 0$ $ax^2+bx+c=0$ $\Rightarrow$ $=>$ standard equation

Comparing ( $1$ $1$ ) with standard equation

$a = 20, b = 1, c = - 13$ $a=20,b=1,c=-13$

Using quadratic formula:

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

$\Rightarrow \frac{- 1 \pm \sqrt{1^{2} - 4 (20 \times - 13)}}{2 (20)}$ $=>(-1+-sqrt(1^2-4(20×-13)))/(2(20))$

$\Rightarrow \frac{- 1 \pm \sqrt{1 - 4 (- 260)}}{40}$ $=>(-1+-sqrt(1-4(-260)))/40$

$\Rightarrow \frac{- 1 \pm \sqrt{1 + 1040}}{40}$ $=>(-1+-sqrt(1+1040))/40$

$\Rightarrow \frac{- 1 \pm \sqrt{1041}}{40}$ $=>(-1+-sqrt(1041))/40$

$\Rightarrow \frac{- 1 \pm 32.26}{40}$ $=>(-1+-32.26)/40$

$x \approx \frac{- 1 + 32.26}{40}, \frac{- 1 - 32.26}{40}$ $x~~(-1+32.26)/40,(-1-32.26)/40$

$x \approx \frac{31.26}{40}, - \frac{33.26}{40}$ $x~~31.26/40,-33.26/40$

$x \approx 0.78, 0.83$ $x~~0.78,0.83$

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How do you solve $20 x^{2} = 13 - x$ $20x^2 = 13 - x$ ?

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