How do you solve the equation $2x^2+3x-5=0$ by completing the square?

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Answer 1 · 2017-02-03T11:22:33

The solutions are $S={1,-5/2}$

Let's complete the square

$2x^2+3x-5=0$

$2(x^2+3/2x)-5=0$

$2(x^2+3/2x+9/16)-5-9/8=0$

$2(x+3/4)^2-49/8=0$

$2(x+3/4)^2=49/16$

$(x+3/4)^2=49/16$

Taking the square roots

$(x+3/4)=+-7/4$

$x=-3/4+-7/4$

$x_1=-10/4=-5/2$

$x_2=4/4=1$

How do you solve the equation 2x^2+3x-5=02x^2+3x-5=0 by completing the square?