Respuesta :

gmany
[tex]5x=6x^2-3\ \ \ |-5x\\\\6x^2-5x-3=0\\\\a=6;\ b=-5;\ c=-3\\\\b^2-4ac=(-5)^2-4\cdot6\cdot(-3)=25+72=97\\\\x_1=\dfrac{-b-\sqrt{b^2-4ac}}{2a};\ x_2=\dfrac{-b+\sqrt{b^2-4ac}}{2a}\\\\x_1=\dfrac{-(-5)-\sqrt{97}}{2\cdot6}=\dfrac{5-\sqrt{97}}{12}\\\\x_2=\dfrac{-(-5)+\sqrt{97}}{2\cdot6}=\dfrac{5+\sqrt{97}}{12}[/tex]

Answer:

[tex]x=\frac{5+\sqrt{97}}{12}[/tex]  and [tex]x=\frac{5-\sqrt{97}}{12}[/tex]

Step-by-step explanation:

Using the quadratic formula to solve [tex]5x=6x^2-3[/tex]

To apply quadratic formula we make 0 on one side of the equation

[tex]5x=6x^2-3[/tex]

Subtract 5x on both sides

[tex]0=6x^2-5x-3[/tex]

Now apply quadratic formula. a=6, b=-5 and c=-3

[tex]x=\frac{-b+-\sqrt{b^2-4ac}}{2a}[/tex]

Plug in all the values in the formula

[tex]x=\frac{5+-\sqrt{(-5)^2-4(6)(-3)}}{2(6)}[/tex]

[tex]x=\frac{5+-\sqrt{97}}{12}[/tex]

[tex]x=\frac{5+\sqrt{97}}{12}[/tex]  and [tex]x=\frac{5-\sqrt{97}}{12}[/tex]

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