# (t + 2)^2/3 + 3(t + 2)^1/3 - 10 =

$\frac{{\left(t+2\right)}^{2}}{3}+3\frac{{\left(t+2\right)}^{1}}{3}-10=0$
how do i solve?
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Dora

Given equation is: $\frac{{\left(t+2\right)}^{2}}{3}+3\left(t+2\right)\frac{1}{3}-10=0\dots \left(1\right)$
Let $\frac{{\left(t+2\right)}^{1}}{3}=x$
$\frac{{\left(t+2\right)}^{1}}{3}\left(t+2\right)13+3\left(t+2\right)13-10=0$
$x\ast x+3\ast x-10=0$
${x}^{2}+3x-10=0\dots \left(2\right)$
Solving the quadratic equation (2) ${x}^{2}+3x-10=0$
${x}^{2}+5x-2x-10$
$x\left(x+5\right)-2\left(x+5\right)=0$
$\left(x+5\right)\left(x-2\right)=0$
x=−5 and x=2
Putting the value of x : $\frac{{\left(t+2\right)}^{1}}{3}=x$
$\frac{{\left(t+2\right)}^{1}}{3}=2$
cubing both the sides $\left(t+2\right)={2}^{3}$
$\left(t+2\right)=8$
$t=8-2=6$
$t=6$
$\frac{{\left(t+2\right)}^{1}}{3}=-5$
$\left(t+2\right)=-125$
$t=-125-2=-127$
Thus the value of t are 6 and -127

Jeffrey Jordon