How do you solve $2 x^{2} + 6 x - 3 = 0$ $2x^2+6x-3=0$ by completing the square?

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How do you solve $2 x^{2} + 6 x - 3 = 0$ $2x^2+6x-3=0$ by completing the square?

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Answer 1 · 2015-04-29T16:26:34

In this way:

$2 x^{2} + 6 x - 3 = 0 \Rightarrow 2 (x^{2} + 3 x) - 3 = 0 \Rightarrow$ $2x^2+6x-3=0rArr2(x^2+3x)-3=0rArr$

$2 (x^{2} + 3 x + \frac{9}{4} - \frac{9}{4}) - 3 = 0 \Rightarrow$ $2(x^2+3x+9/4-9/4)-3=0rArr$

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

$2 {(x + \frac{3}{2})}^{2} - \frac{9}{2} - 3 = 0 \Rightarrow$ $2(x+3/2)^2-9/2-3=0rArr$

$2 {(x + \frac{3}{2})}^{2} = \frac{9 + 6}{2} \Rightarrow {(x + \frac{3}{2})}^{2} = \frac{15}{4} \Rightarrow$ $2(x+3/2)^2=(9+6)/2rArr(x+3/2)^2=15/4rArr$

$x + \frac{3}{2} = \pm \frac{\sqrt{15}}{2} \Rightarrow x = - \frac{3}{2} \pm \frac{\sqrt{15}}{2}$ $x+3/2=+-sqrt15/2rArrx=-3/2+-sqrt15/2$ .

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How do you solve $2 x^{2} + 6 x - 3 = 0$ $2x^2+6x-3=0$ by completing the square?

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How do you solve $2 x^{2} + 6 x - 3 = 0$ $2x^2+6x-3=0$ by completing the square?