How do you solve by completing the square $2 x^{2} - 6 x - 20 = 0$ $2x^2-6x-20=0$ ?

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Answer 1 · 2018-03-23T18:28:11

$x_{1} = - 2$ $x_1=-2$ and $x_{2} = 5$ $x_2=5$

$2 x^{2} - 6 x - 20 = 0$ $2x^2-6x-20=0$

$2 \cdot (x^{2} - 3 x - 10) = 0$ $2*(x^2-3x-10)=0$

$2 \cdot (x^{2} - 3 x + \frac{9}{4} - \frac{49}{4}) = 0$ $2*(x^2-3x+9/4-49/4)=0$

$2 \cdot ({(x - \frac{3}{2})}^{2} - {(\frac{7}{2})}^{2}) = 0$ $2*((x-3/2)^2-(7/2)^2)=0$

$2 (x - \frac{3}{2} + \frac{7}{2}) (x - \frac{3}{2} - \frac{7}{2}) = 0$ $2(x-3/2+7/2)(x-3/2-7/2)=0$

$2 (x + 2) (x - 5) = 0$ $2(x+2)(x-5)=0$

Thus, $x_{1} = - 2$ $x_1=-2$ and $x_{2} = 5$ $x_2=5$

How do you solve by completing the square 2x2−6x−20=02x^2-6x-20=0?