$\text{The quadratic formula for a quadratic equation}\ ax^2+bx+c=0\\\\\text{determinant:}\ \Delta=b^2-4ac\\\\\text{If}\ \Delta0,\ \text{then the equation has two different solutions}\ x=\dfrac{-b\pm\sqrt\Delta}{2a}.$

$\text{We have:}\\\\3x^2+8x-4=0\\\\a=3,\ b=8,\ c=-4\\\\\Delta=8^2-4(3)(-4)=64+48=112>0\\\sqrt\Delta=\sqrt{112}=\sqrt{(16)(7)}=\sqrt{16}\cdot\sqrt7=4\sqrt7\\\\x=\dfrac{-8\pm4\sqrt7}{2(3)}=\dfrac{-8\pm4\sqrt7}{6}=\dfrac{2(-4\pm2\sqrt7)}{6}=\dfrac{-4\pm2\sqrt7}{3}$

3 0

3 years ago

Plz help!!!!!!!!!!!!!!!!!!

e-lub [12.9K]

The original triangle had side lengths of 12. The perimeter was 36

Then if you divide each side by 4, each side would be 3 and the new triangle's perimeter would be 9. The difference of them would be 36-9 which is 27. The answer is 27. Hope that helped

4 0

3 years ago

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