# 310=3x^{2}−x

$310=3{x}^{2}-x$
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Liyana Mansell

To solve $310=3{x}^{2}-x$, we first need to move all the terms to one side to write the equation in standard form. Subtracting 310 on both sides then gives $0=3{x}^{2}-x-310$.
To factor a quadratic of the form $a{x}^{2}+bx+c$, you need to find two numbers, m and n, such that $mn=ac$ and $m+n=b$. You can then rewrite the middle term as bx=mx+nx and factor by grouping.
For $3{x}^{2}-x-310$, a=3, b=−1, and  $c=-310$ so $mn=ac=3\left(-310\right)=-930$ and $m+n=b=-1$. Therefore $m=-31$ and $n=30$ since $\left(-31\right)\left(30\right)=-930$ and $-31+30=-1$. You can then rewrite the middle term as $-x=-31x+30x$ and then factor by grouping:
$3{x}^{2}-x-310=0$
$3{x}^{2}-31x+30x-310=0$  Rewrite the middle term.
$\left(3{x}^{2}\left\{2\right\}-31x\right)+\left(30x-310\right)=0$ Group each pair of terms.
$x\left(3x-31\right)+10\left(3x-31\right)=0$  Factor out the GCF of each pair.
$\left(3x-31\right)\left(x+10\right)=0$ Factor out $3x-31$.
Now that we have the equation factored, we can use the Zero Product Property to solve for x. The Zero Product Property states that if $ab=0$, then $a=0$ or $b=0$. Therefore, if $\left(3x-31\right)\left(x+10\right)=0$, then . Solving each of these for x gives:
$3x-31=0$

$x+10=0$
$3x=31$

$x=-10$
$x=\frac{31}{3}$
The solutions of the equation are then $x=\frac{31}{3}$ and $x=-10$