Respuesta :
Solve :
[tex]3x^2+4x=2[/tex]so, 3x^2 + 4x - 2 = 0
using the formula:
[tex]x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]so, a = 3 , b = 4 and c = -2
so,
[tex]x=\frac{-4\pm\sqrt{4^2-4\cdot3\cdot(-2)}}{2\cdot3}=\frac{-4\pm\sqrt{16+24}}{6}=\frac{-4\pm\sqrt{40}}{6}[/tex][tex]x_1=\frac{-4+\sqrt{40}}{6}=0.387[/tex]And
[tex]x_2=\frac{-4-\sqrt{40}}{6}=-1.721[/tex]so, x = 0.387 or -1.721
The answer is rounded to 3 decimals.
make it without simplifications
[tex]\sqrt{40}=\sqrt{4\cdot10}=2\sqrt{10}[/tex][tex]x\text{ = }\frac{-4\pm2\sqrt{10}}{6}=\frac{2\text{ (-2}\pm\sqrt{10})}{2\cdot3}=\frac{-2\pm\sqrt{10}}{3}[/tex][tex]x\text{ = }\frac{-2+\sqrt{10}}{3}\text{ OR }x\text{ =}\frac{-2-\sqrt{10}}{3}[/tex]
