How do you find the real or imaginary solutions of the equation $6 x^{2} + 13 x - 5 = 0$ $6x^2+13x-5=0$ ?

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Answer 1 · 2016-11-28T10:56:34

The solutions are $S = {\frac{1}{3}, - \frac{5}{2}}$ $S={1/3,-5/2}$

The simultaneous equations $a x^{2} + b x + c = 0$ $ax^2+bx+c=0$

Our equation is $6 x^{2} + 13 x - 5 = 0$ $6x^2+13x-5=0$

We calculate the determinant,

$Delta=b^2-4ac$

$Delta=13^2-4*6*-5=289$

$Delta>0$ , so we have 2 real roots

So,

$x=(-b+-sqrtDelta)/(2a)$

$=(-13+-sqrt289)/12$

$=(-13+-17)/12$

So, $x_1=-30/12=-5/2$

and $x_2=4/12=1/3$

How do you find the real or imaginary solutions of the equation 6x^2+13x-5=06x2+13x−5=06x^2+13x-5=0?