Nayraidn
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what are the zeros of -3x^2+10x+12

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Konrad509idn
[tex] -3x^2+10x+12=0\\ \Delta=10^2-4\cdot(-3)\cdot12=100+144=244\\ \sqrt{\Delta}=\sqrt{244}=2\sqrt{61}\\ x_1=\dfrac{-10-2\sqrt{61}}{2\cdot(-3)}=\dfrac{-10-2\sqrt{61}}{-6}=\dfrac{5+\sqrt{61}}{3}\\ x_2=\dfrac{-10+2\sqrt{61}}{2\cdot(-3)}=\dfrac{-10+2\sqrt{61}}{-6}=\dfrac{5-\sqrt{61}}{3}\\[/tex]
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