Example

3(r+2s)=2t-4
a⋅(b-2)=3b
3(r+2s)=2t-4
a⋅(b-2)=3b

Question

$${ x }^{ 2 } +(x+1) { \left( \right) }^{ 2 } =365$$

Answer

$$x=(-^2+sqrt(^4-4*^2+1460))/2,(-^2-sqrt(^4-4*^2+1460))/2$$

Solution


Expand.
\[{x}^{2}+{}^{2}x+{}^{2}=365\]
Move all terms to one side.
\[{x}^{2}+{}^{2}x+{}^{2}-365=0\]
Use the Quadratic Formula.
\[x=\frac{-{}^{2}+\sqrt{{}^{4}-4{}^{2}+1460}}{2},\frac{-{}^{2}-\sqrt{{}^{4}-4{}^{2}+1460}}{2}\]
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