How do you solve using the quadratic formula method $2x^2 + 3x – 1 = 0$ ?

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Answer 1 · 2016-03-07T14:39:51

$x=(-3+sqrt17)/4$ or $x=(-3-sqrt17)/4$

Quadratic formula which gives solution of $ax^2+bx+c=0$
is $x=(-b+-sqrt(b^2-4ac))/(2a)$ .

In the equation $2x^2+3x–1=0$ , $a=2$ , $b=3$ and $c=-1$ and hence solution is

$x=(-3+-sqrt(3^2-4xx2xx(-1)))/(2xx2)$

or $x=(-3+-sqrt(9+8))/(2xx2)=-(3+-sqrt17)/4$

i.e. $x=(-3+sqrt17)/4$ or $x=(-3-sqrt17)/4$

How do you solve using the quadratic formula method 2x^2 + 3x – 1 = 02x^2 + 3x – 1 = 0?