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Example
3(r+2s)=2t-4
a⋅(b-2)=3b
3(r+2s)=2t-4
a⋅(b-2)=3b
Question
$$5 { x }^{ 2 } +8x+1=0$$
Answer
x=-(4-sqrt(11))/5,-(4+sqrt(11))/5
Solution
Use the Quadratic Formula.
In general, given \(a{x}^{2}+bx+c=0\), there exists two solutions where:
\[x=\frac{-b+\sqrt{{b}^{2}-4ac}}{2a},\frac{-b-\sqrt{{b}^{2}-4ac}}{2a}\]
In this case, \(a=5\), \(b=8\) and \(c=1\).
\[{x}^{}=\frac{-8+\sqrt{{8}^{2}-4\times 5}}{2\times 5},\frac{-8-\sqrt{{8}^{2}-4\times 5}}{2\times 5}\]
Simplify.
\[x=\frac{-8+2\sqrt{11}}{10},\frac{-8-2\sqrt{11}}{10}\]
\[x=\frac{-8+2\sqrt{11}}{10},\frac{-8-2\sqrt{11}}{10}\]
Simplify solutions.
\[x=-\frac{4-\sqrt{11}}{5},-\frac{4+\sqrt{11}}{5}\]
Decimal Form: -0.136675, -1.463325
x=-(4-sqrt(11))/5,-(4+sqrt(11))/5
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