Radical equation 2x^2-5x-3=0

radical equation need help
2x^2-5x-3=0

First you have to factor 2x^2-5x-3
You should get (2x+1)(x-3)=0
Then you set (2x+1) equal to zero, and (x-3) equal to zero. (Because any thing times zero equals zero.)
Solve for x.

Answers: x={-1/2,3}

If you substitute both answers in the original equation you should get zero equals zero both times.

You can solve this problem by completing the square or by using the quadratic equation. I will solve by the using the quadradic equation since it is a quadradic formula.

The equation of a quadratic formula= (-b+- sqrt(-b-4ac))/2a

a=2, b=-5, c=-3 so...
(-( -5)+- sqrt(-5)^2 -4(2)(-3))/2(2)

so... (5+- sqrt(25+24))/4 = (5+- sqrt(49))/4

so... (5+7)/4 = 3 or (5-7)/4 = -1/2

So your answers are 3 and -1/2

(2x + 1)(x - 3) = 0

True when 2x + 1 = 0 or when x - 3 = 0

x = -1/2, 3

{x = -(1/2)},

{x = 3}

factorise 2x^2-6x+x-3=0
2x(x-3)+(x-3)=(x-3)(2x+1)=0 so x=3 or x=-1/2

x=3, -1/2