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Solve 2x^2 – 13x + 14 = 0 to the nearest tenth. Submit Answer

User MosheK
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1 Answer

6 votes

Answer:

x = 5.1 and x = 1.4

Step-by-step explanation:

When we have an equation with the form:

ax²+bx+c=0

The solutions of the equation can be calculated using the following equation:


\begin{gathered} x=\frac{-b+\sqrt[]{b^2-4ac}}{2a} \\ x=\frac{-b-\sqrt[]{b^2-4ac}}{2a} \end{gathered}

So, our equation is 2x²-13x+14=0 where a is 2, b is - 13 and c is 14. Then, the solutions are:


\begin{gathered} x=\frac{-(-13)+\sqrt[]{(-13)^2-4(2)(14)}}{2(2)}=\frac{13+\sqrt[]{57}}{4}=5.1 \\ x=\frac{-(-13)-\sqrt[]{(-13)^2-4(2)(14)}}{2(2)}=\frac{13-\sqrt[]{57}}{4}=1.4 \end{gathered}

Therefore, the solutions to the nearest tenth are 5.1 and 1.4

User Dhvanil
by
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