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Basic Math
Arithmetic
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Integers & Absolute Values
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Example
3(r+2s)=2t-4
a⋅(b-2)=3b
3(r+2s)=2t-4
a⋅(b-2)=3b
Question
$${ x }^{ 2 } +28x+105=0$$
Answer
x=-14+sqrt(91),-14-sqrt(91)
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, \(a=1\), \(b=28\) and \(c=105\).
\[{x}^{}=\frac{-28+\sqrt{{28}^{2}-4\times 105}}{2},\frac{-28-\sqrt{{28}^{2}-4\times 105}}{2}\]
Simplify.
\[x=\frac{-28+2\sqrt{91}}{2},\frac{-28-2\sqrt{91}}{2}\]
\[x=\frac{-28+2\sqrt{91}}{2},\frac{-28-2\sqrt{91}}{2}\]
Simplify solutions.
\[x=-14+\sqrt{91},-14-\sqrt{91}\]
Decimal Form: -4.460608, -23.539392
x=-14+sqrt(91),-14-sqrt(91)
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