Question

Using the quadratic formula to solve 5x=6x^2-3 , what are the values of x

Answer 1

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

Answer 2

Answer:

$x=\frac{5+\sqrt{97}}{12}$  and $x=\frac{5-\sqrt{97}}{12}$

Step-by-step explanation:

Using the quadratic formula to solve $5x=6x^2-3$

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

$5x=6x^2-3$

Subtract 5x on both sides

$0=6x^2-5x-3$

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

$x=\frac{-b+-\sqrt{b^2-4ac}}{2a}$

Plug in all the values in the formula

$x=\frac{5+-\sqrt{(-5)^2-4(6)(-3)}}{2(6)}$

$x=\frac{5+-\sqrt{97}}{12}$

$x=\frac{5+\sqrt{97}}{12}$  and $x=\frac{5-\sqrt{97}}{12}$