Solve 10^2-E^2-4Fquad=quadsqrt{8^2+10^2-4} | Microsoft Math Solver

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Step by step: \begin{array}{rcl} && x + jy = (a + jb)^{c+jd} \\ &=& (re^{j\theta})^{c+jd} \\ &=& r^{c+jd} (e^{j\theta})^{c+jd} \\ &=& r^cr^{jd} e^{j\theta c}e^{-\theta d} \\ &=& r^c e^{\displaystyle\ln r^{jd}} e^{j\theta c}e^{-\theta d} \\ &=& r^c e^{-\theta d} e^{\displaystyle jd \ln r} e^{\displaystyle j\theta c} \\ &=& r^c e^{-\theta d} e^{\displaystyle j(d \ln r + \theta c)} \\ &=& r^c e^{-\theta d} [\cos(d \ln r + \theta c) + j\sin(d \ln r + \theta c)] \\ &=& [r^c e^{-\theta d} \cos(d \ln r + \theta c)] + j[r^c e^{-\theta d}\sin(d \ln r + \theta c)] \\ &=& \left[(a^2+b^2)^{c/2} e^{-\theta d} \cos\left(d \ln(\sqrt{a^2+b^2}) + \theta c\right)\right] + \\ && \left[(a^2+b^2)^{c/2} e^{-\theta d}\sin\left(d \ln((a^2+b^2)^{1/2}) + \theta c\right)\right]j \\ &=& \left[(a^2+b^2)^{c/2} e^{-\theta d} \cos\left(\frac{d}{2} \ln(a^2+b^2) + \theta c\right)\right] + \\ && \left[(a^2+b^2)^{c/2} e^{-\theta d}\sin\left(\frac{d}{2} \ln(a^2+b^2) + \theta c\right)\right]j \\ &=& \begin{cases} \left[(a^2+b^2)^{c/2} e^{-d\arctan(\frac{b}{a})} \cos\left(\frac{d}{2} \ln(a^2+b^2) + c\,\arctan(\frac{b}{a})\right)\right] + &\\ \left[(a^2+b^2)^{c/2} e^{-d\arctan(\frac{b}{a})} \sin\left(\frac{d}{2} \ln(a^2+b^2) + c\,\arctan(\frac{b}{a})\right)\right]j & \text{if} \; a\!>\!0 \\ \left[(a^2+b^2)^{c/2} e^{-d\arctan(\frac{b}{a}) - \pi d} \cos\left(\frac{d}{2} \ln(a^2+b^2) + c\,\arctan(\frac{b}{a}) + \pi c\right)\right] + & \\ \left[(a^2+b^2)^{c/2} e^{-d\arctan(\frac{b}{a}) - \pi d} \sin\left(\frac{d}{2} \ln(a^2+b^2) + c\,\arctan(\frac{b}{a}) + \pi c\right)\right]j & \text{if} \; a\!<\!0 \; \text{and} \; b \!\ge\! 0 \\ \left[(a^2+b^2)^{c/2} e^{-d\arctan(\frac{b}{a}) + \pi d} \cos\left(\frac{d}{2} \ln(a^2+b^2) + c\,\arctan(\frac{b}{a}) - \pi c\right)\right] + & \\ \left[(a^2+b^2)^{c/2} e^{-d\arctan(\frac{b}{a}) + \pi d} \sin\left(\frac{d}{2} \ln(a^2+b^2) + c\,\arctan(\frac{b}{a}) - \pi c\right)\right]j & \text{if} \; a\!<\!0 \; \text{and} \; b\!<\!0 \\ \left[(a^2+b^2)^{c/2} e^{-\frac{\pi d}{2}} \cos\left(\frac{d}{2} \ln(a^2+b^2) + \frac{\pi c}{2}\right)\right] + & \\ \left[(a^2+b^2)^{c/2} e^{-\frac{\pi d}{2}} \sin\left(\frac{d}{2} \ln(a^2+b^2) + \frac{\pi c}{2}\right)\right]j & \text{if} \; a\!=\!0 \; \text{and} \; b\!>\!0 \\ \left[(a^2+b^2)^{c/2} e^{+\frac{\pi d}{2}} \cos\left(\frac{d}{2} \ln(a^2+b^2) - \frac{\pi c}{2}\right)\right] + & \\ \left[(a^2+b^2)^{c/2} e^{+\frac{\pi d}{2}} \sin\left(\frac{d}{2} \ln(a^2+b^2) - \frac{\pi c}{2}\right)\right]j & \text{if} \; a\!=\!0 \; \text{and} \; b\!<\!0 \\ 0 & \text{if} \; a\!=\!0 \; \text{and} \; b\!=\!0 \\ \end{cases} \end{array} ...

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100-E^{2}-4F=\sqrt{8^{2}+10^{2}-4}

Calculate 10 to the power of 2 and get 100.

100-E^{2}-4F=\sqrt{64+10^{2}-4}

Calculate 8 to the power of 2 and get 64.

100-E^{2}-4F=\sqrt{64+100-4}

Calculate 10 to the power of 2 and get 100.

100-E^{2}-4F=\sqrt{164-4}

Add 64 and 100 to get 164.

100-E^{2}-4F=\sqrt{160}

Subtract 4 from 164 to get 160.

100-E^{2}-4F=4\sqrt{10}

Factor 160=4^{2}\times 10. Rewrite the square root of the product \sqrt{4^{2}\times 10} as the product of square roots \sqrt{4^{2}}\sqrt{10}. Take the square root of 4^{2}.

-E^{2}-4F=4\sqrt{10}-100

Subtract 100 from both sides.

-4F=4\sqrt{10}-100+E^{2}

Add E^{2} to both sides.

-4F=E^{2}+4\sqrt{10}-100

The equation is in standard form.

\frac{-4F}{-4}=\frac{E^{2}+4\sqrt{10}-100}{-4}

Divide both sides by -4.

F=\frac{E^{2}+4\sqrt{10}-100}{-4}

Dividing by -4 undoes the multiplication by -4.

F=-\frac{E^{2}}{4}+25-\sqrt{10}

Divide 4\sqrt{10}-100+E^{2} by -4.

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