How do you complete the square to solve $2 x^{2} + 6 x = 5$ $2x^2 + 6x =5$ ?

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Answer 1 · 2015-05-24T11:31:02

$2 x^{2} + 6 x = 5$ $2x^2+6x = 5$

Extract the common constant factor on the left side
$2 (x^{2} + 3 x) = 5$ $2(x^2+3x) = 5$

Complete the square
$2 (x^{2} + 3 x + {(\frac{3}{2})}^{2}) = 5 + \frac{9}{2}$ $2(x^2+3x+(3/2)^2) = 5 + 9/2$

Re-write as a square equal to a constant
${(x + \frac{3}{2})}^{2} = \frac{19}{4}$ $(x+3/2)^2 = 19/4$

Take the square root of both sides
$x + \frac{3}{2} = \pm \frac{\sqrt{19}}{2}$ $x+3/2 = +-sqrt(19)/2$

Isolate $x$ $x$ for the final solution
$x = \frac{- 3 \pm \sqrt{19}}{2}$ $x = (-3+-sqrt(19))/2$

How do you complete the square to solve 2x2+6x=5 2x^2 + 6x =5?