Question

`{40a}^(2) + 20a - 30 = ( - 1)( {15a}^(3) + 30)`
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Answer 1

`\bf 40a^2+20a-30=(-1)(15a^3+30)\implies 40a^2+20a~~\begin{matrix} -30 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~=-15a^3~~\begin{matrix} -30 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix} \\\\\\ 40a^2+20a=-15a^3\implies 15a^3+40a^2+20a=0\implies 5a(3a^2+8a+4)=0 \\\\\\ 5a(3a+2)(a+2)=0\implies a= \begin{cases} 0\\ -(2)/(3)\\ -2 \end{cases}`