Question

x(4x-1)=3(x+1) ?????​

Answer 1

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$\sf x\cdot(4x-1)=3\cdot(x+1)\\\sf 4x^2-x=3x+3\\\sf 4x^2-x-3x-3=0\\\sf 4x^2-4x-3=0\\\sf a=4~~b=-4~c=-3\\\sf\Delta=b^2-4ac\\\sf\Delta=(-4)^2-4\cdot4\cdot(-3)\\\sf\Delta=16+48\\\sf\Delta=64\\\sf x=\dfrac{-b\pm\sqrt{\Delta}}{2a}\\\sf x=\dfrac{-(-4)\pm\sqrt{64}}{2\cdot4}\\\sf x=\dfrac{4\pm8}{8}\begin{cases}\sf x_1=\dfrac{4+8}{8}=\dfrac{12\div4}{8\div4}=\dfrac{3}{2}\\\sf x_2=\dfrac{4-8}{8}=-\dfrac{4\div4}{8\div4}=-\dfrac{1}{2}\end{cases}$

$\huge\boxed{\boxed{\boxed{\boxed{\sf S=\bigg\{-\dfrac{1}{2},\dfrac{3}{2}\bigg\}}}}}\checkmark$