Respuesta :

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]

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