Question

solve for x: 6x^2 - 13x + 5 = 0
x = -\frac{1}{3} or x = \frac{5}{2}
x = \frac{1}{3} or x = -\frac{5}{2}
x = \frac{1}{2} or x = \frac{5}{3}
x = -\frac{1}{2} or x = -\frac{5}{3}

Explanation:

Step1: Factor the quadratic equation

For the quadratic equation $6x^{2}-13x + 5=0$, we need to find two numbers that multiply to $6\times5 = 30$ and add up to $- 13$. The numbers are $-10$ and $-3$. So we can rewrite the middle - term: $6x^{2}-10x-3x + 5 = 0$. Then factor by grouping: $2x(3x - 5)-(3x - 5)=0$, which gives $(2x - 1)(3x - 5)=0$.

Step2: Set each factor equal to zero

If $(2x - 1)(3x - 5)=0$, then either $2x-1 = 0$ or $3x - 5=0$.
For $2x-1 = 0$, we solve for $x$:
$2x=1$, so $x=\frac{1}{2}$.
For $3x - 5=0$, we solve for $x$:
$3x=5$, so $x=\frac{5}{3}$.

Answer:

$x=\frac{1}{2}$ or $x=\frac{5}{3}$