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19 votes
19 votes
So i have to factor x(3x+10)=77

User Anton Geraschenko
by
2.8k points

1 Answer

4 votes
4 votes

SOLUTION

Given the equation as seen below, we can use the following steps to get the factors


x(3x+10)=77

Step 1: Remove the bracket by multiplying the value outside the bracket with the one inside the bracket using the distributive law. We have:


\begin{gathered} x(3x)+x(10)=77 \\ 3x^2_{}+10x=77 \\ 3x^2+10x-77=0 \end{gathered}

Step 2: Now that we have a quadratic equation, we solve for x using the quadratic formula:


\begin{gathered} 3x^2+10x-77=0 \\ u\sin g\text{ the form }ax^2+bx+c=0 \\ a=3,b=10,c=-77 \\ x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ =\frac{-10\pm\sqrt[]{10^2-4(3)(-77)}}{2(3)} \\ =\frac{-10\pm\sqrt[]{100+924}}{6} \\ =\frac{-10\pm\sqrt[]{1024}}{6} \\ =(-10+32)/(6)\text{ or }(-10-32)/(6) \\ (22)/(6)\text{ or -}(42)/(6) \\ =(11)/(3)\text{ or }-7 \end{gathered}

Hence, it can be seen from above that the factors will be -7 or 11/3.

User Stevenelberger
by
3.3k points
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