914.06 = 50/(1+Y)^1 + 1050(1+Y)^2
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You asked:
Solve the equation
914.06 = 50 ( 1 + Y ) 1 + 1050 ( 1 + Y ) 2 914.06 = \frac{50}{{\left( 1 + Y \right)}^{1}} + 1050 {\left( 1 + Y \right)}^{2} 914.06 = ( 1 + Y ) 1 50 + 1050 ( 1 + Y ) 2 for the variable
Y Y Y .
MathBot Answer:
The 3 solutions to the equation are:
Y = − 1 − 6529 cos ( arctan ( 13321135270561 562500 ) 3 ) 75 ≈ − 1.9592575 Y = − 1 − 19587 sin ( arctan ( 13321135270561 562500 ) 3 ) 300 − 6529 re ( 1 ( − 1 2 − 3 i 2 ) 9 14 + 13321135270561 i 875000 3 ) 7500 + 6529 cos ( arctan ( 13321135270561 562500 ) 3 ) 300 + i ( − 6529 im ( 1 ( − 1 2 − 3 i 2 ) 9 14 + 13321135270561 i 875000 3 ) 7500 + 6529 sin ( arctan ( 13321135270561 562500 ) 3 ) 300 + 19587 cos ( arctan ( 13321135270561 562500 ) 3 ) 300 ) ≈ − 0.94510901 + 2.0 ⋅ 1 0 − 143 i Y = − 1 + 19587 sin ( arctan ( 13321135270561 562500 ) 3 ) 300 + 6529 cos ( arctan ( 13321135270561 562500 ) 3 ) 300 − 6529 re ( 1 ( − 1 2 + 3 i 2 ) 9 14 + 13321135270561 i 875000 3 ) 7500 + i ( − 19587 cos ( arctan ( 13321135270561 562500 ) 3 ) 300 + 6529 sin ( arctan ( 13321135270561 562500 ) 3 ) 300 − 6529 im ( 1 ( − 1 2 + 3 i 2 ) 9 14 + 13321135270561 i 875000 3 ) 7500 ) ≈ − 0.095633513 − 2.0 ⋅ 1 0 − 142 i \begin{aligned}Y &= -1 - \frac{\sqrt{6529} \cos{\left(\frac{\arctan{\left(\frac{\sqrt{13321135270561}}{562500} \right)}}{3} \right)}}{75} \approx -1.9592575\\Y &= -1 - \frac{\sqrt{19587} \sin{\left(\frac{\arctan{\left(\frac{\sqrt{13321135270561}}{562500} \right)}}{3} \right)}}{300} - \frac{6529 \operatorname{re}{\left(\frac{1}{\left(- \frac{1}{2} - \frac{\sqrt{3} i}{2}\right) \sqrt[3]{\frac{9}{14} + \frac{\sqrt{13321135270561} i}{875000}}}\right)}}{7500} + \frac{\sqrt{6529} \cos{\left(\frac{\arctan{\left(\frac{\sqrt{13321135270561}}{562500} \right)}}{3} \right)}}{300} + i \left(- \frac{6529 \operatorname{im}{\left(\frac{1}{\left(- \frac{1}{2} - \frac{\sqrt{3} i}{2}\right) \sqrt[3]{\frac{9}{14} + \frac{\sqrt{13321135270561} i}{875000}}}\right)}}{7500} + \frac{\sqrt{6529} \sin{\left(\frac{\arctan{\left(\frac{\sqrt{13321135270561}}{562500} \right)}}{3} \right)}}{300} + \frac{\sqrt{19587} \cos{\left(\frac{\arctan{\left(\frac{\sqrt{13321135270561}}{562500} \right)}}{3} \right)}}{300}\right) \approx -0.94510901 + 2.0 \cdot 10^{-143} i\\Y &= -1 + \frac{\sqrt{19587} \sin{\left(\frac{\arctan{\left(\frac{\sqrt{13321135270561}}{562500} \right)}}{3} \right)}}{300} + \frac{\sqrt{6529} \cos{\left(\frac{\arctan{\left(\frac{\sqrt{13321135270561}}{562500} \right)}}{3} \right)}}{300} - \frac{6529 \operatorname{re}{\left(\frac{1}{\left(- \frac{1}{2} + \frac{\sqrt{3} i}{2}\right) \sqrt[3]{\frac{9}{14} + \frac{\sqrt{13321135270561} i}{875000}}}\right)}}{7500} + i \left(- \frac{\sqrt{19587} \cos{\left(\frac{\arctan{\left(\frac{\sqrt{13321135270561}}{562500} \right)}}{3} \right)}}{300} + \frac{\sqrt{6529} \sin{\left(\frac{\arctan{\left(\frac{\sqrt{13321135270561}}{562500} \right)}}{3} \right)}}{300} - \frac{6529 \operatorname{im}{\left(\frac{1}{\left(- \frac{1}{2} + \frac{\sqrt{3} i}{2}\right) \sqrt[3]{\frac{9}{14} + \frac{\sqrt{13321135270561} i}{875000}}}\right)}}{7500}\right) \approx -0.095633513 -2.0 \cdot 10^{-142} i\end{aligned} Y Y Y = − 1 − 75 6529 cos ( 3 a r c t a n ( 562500 13321135270561 ) ) ≈ − 1.9592575 = − 1 − 300 19587 sin ( 3 a r c t a n ( 562500 13321135270561 ) ) − 7500 6529 re ( ( − 2 1 − 2 3 i ) 3 14 9 + 875000 13321135270561 i 1 ) + 300 6529 cos ( 3 a r c t a n ( 562500 13321135270561 ) ) + i − 7500 6529 im ( ( − 2 1 − 2 3 i ) 3 14 9 + 875000 13321135270561 i 1 ) + 300 6529 sin ( 3 a r c t a n ( 562500 13321135270561 ) ) + 300 19587 cos ( 3 a r c t a n ( 562500 13321135270561 ) ) ≈ − 0.94510901 + 2.0 ⋅ 1 0 − 143 i = − 1 + 300 19587 sin ( 3 a r c t a n ( 562500 13321135270561 ) ) + 300 6529 cos ( 3 a r c t a n ( 562500 13321135270561 ) ) − 7500 6529 re ( ( − 2 1 + 2 3 i ) 3 14 9 + 875000 13321135270561 i 1 ) + i − 300 19587 cos ( 3 a r c t a n ( 562500 13321135270561 ) ) + 300 6529 sin ( 3 a r c t a n ( 562500 13321135270561 ) ) − 7500 6529 im ( ( − 2 1 + 2 3 i ) 3 14 9 + 875000 13321135270561 i 1 ) ≈ − 0.095633513 − 2.0 ⋅ 1 0 − 142 i
i i i imaginary unit , defined as i 2 = − 1 i^2 = -1 i 2 = − 1
