Question

$$2a^{2}+3\dot{x}+k=0$$

Answer

$$x=(-2*a^2-k)/(3*d*o*t)$$

Solution


Subtract \(2{a}^{2}\) from both sides.
\[3dotx+k=-2{a}^{2}\]
Subtract \(k\) from both sides.
\[3dotx=-2{a}^{2}-k\]
Divide both sides by \(3\).
\[dotx=\frac{-2{a}^{2}-k}{3}\]
Divide both sides by \(d\).
\[otx=\frac{\frac{-2{a}^{2}-k}{3}}{d}\]
Simplify  \(\frac{\frac{-2{a}^{2}-k}{3}}{d}\)  to  \(\frac{-2{a}^{2}-k}{3d}\).
\[otx=\frac{-2{a}^{2}-k}{3d}\]
Divide both sides by \(o\).
\[tx=\frac{\frac{-2{a}^{2}-k}{3d}}{o}\]
Simplify  \(\frac{\frac{-2{a}^{2}-k}{3d}}{o}\)  to  \(\frac{-2{a}^{2}-k}{3do}\).
\[tx=\frac{-2{a}^{2}-k}{3do}\]
Divide both sides by \(t\).
\[x=\frac{\frac{-2{a}^{2}-k}{3do}}{t}\]
Simplify  \(\frac{\frac{-2{a}^{2}-k}{3do}}{t}\)  to  \(\frac{-2{a}^{2}-k}{3dot}\).
\[x=\frac{-2{a}^{2}-k}{3dot}\]