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Mathematics
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1. 3a^3 - 14a^2 - 5a =0 2. 8x + 4 =0 3. y^2 + 9y + 14 =0
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the first one factorize in a(3a^2-14a-5)=0 now from here we have... a=0 or 3a^2-14a-5=0 to solve 3a^2-14a-5=0 you have the quadratic equation... use \[\Large x_{1/2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\] and you'll have two solutions \[\Large a_1=-\frac13 \quad a_2=5 \quad a_3=0 \quad(\text{a_3 we have taken from principal})\] the second task is easier... 8x+4=0 8x=-4 \[\Large x=-\frac{4}{8} \Longrightarrow x=-\frac12\] now regarding question nr. 3 is similar as part two of the first question ... use quadratic formula and you'll have.... \[\LARGE y_1=-7 \quad \quad y_2=-2\]
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