## Big Ideas Math - Algebra 1, A Common Core Curriculum

$x=\frac{3}{2}$ and $x=\frac{2}{3}$
$6x^{2}-13x=-6\implies 6x^{2}-13x+6=0$ Comparing $6x^{2}-13x+6=0$ with $ax^{2}+bx+c=0$, we get $a=6$, $b=-13$ and $c=6$. The quadratic formula is $x=\frac{-b \pm \sqrt {b^{2}-4ac}}{2a}$ Substituting the values, we get $x=\frac{-(-13) \pm \sqrt {(-13)^{2}-4(6)(6)}}{(2)(6)}=\frac{13\pm 5}{12}$ The solutions of the equation are $x=\frac{13+5}{12}=\frac{18}{12}=\frac{3}{2}$ and $x=\frac{13-5}{12}=\frac{8}{12}=\frac{2}{3}$
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