EXPLANATION
Given the expression 2y(x-y) +12=5x , plugging in x=3 into the expression,
2y(3-y) + 12 = 5*3
6y -2y^2 + 12 = 15
Rearranging terms:
[tex]-2y^2+6y+12=15[/tex]Now, we need to apply the quadratic equation:
[tex]\mathrm{For\: a\: quadratic\: equation\: of\: the\: form\: }ax^2+bx+c=0\mathrm{\: the\: solutions\: are\: }[/tex][tex]x_{1,\: 2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex][tex]\mathrm{For\: }\quad a=-2,\: b=6,\: c=-3[/tex][tex]y_{1,2}=\frac{-6\pm\sqrt{6^2-4\left(-2\right)\left(-3\right)}}{2\left(-2\right)}[/tex]Multiplying terms:
[tex]y_{1,2}=\frac{-6\pm\sqrt[]{36-24}}{-4}[/tex]Subtracting numbers:
[tex]y_{1,2}=\frac{-6\pm\sqrt[]{12}}{-4}[/tex]Simplifying:
[tex]\mathrm{The\: solutions\: to\: the\: quadratic\: equation\: are\colon}[/tex][tex]y=\frac{3-\sqrt{3}}{2},\: y=\frac{3+\sqrt{3}}{2}[/tex]