y= -6x² +3x +2
Using the quadratic equation x= [tex] \frac{-b+ \sqrt{b^2-4ac} }{2a} [/tex]
We assign variables a,b,c as follows with the form ax²+bx +c
a= -6
b= 3
c=2
By plugging in the variable in the quadratic equation we can find x
[tex]x= \frac{-b+ \sqrt{b^2-4ac} }{2a} [/tex]
=[tex]\frac{-3+- \sqrt{3^2-4(-6)(2)} }{2(-6)} [/tex]
If you simplify the equation you will have two real solutions for x
[tex] \frac{-3+- \sqrt{9+48} }{-12} [/tex]
[tex] \frac{-3+- \sqrt{57} }{-12}[/tex]
As a square root can be positive or negative, two solutions are possible
Hence,
[tex]x = \frac{-3+\sqrt{57} }{-12}, \frac{-3- \sqrt{57} }{-12}[/tex]
Which can be further simplified to
[tex]x = \frac{\sqrt{57} }{4}, \frac{- \sqrt{57} }{4}[/tex]
TAA DAAA! :)
PS if you rate my answer as "Brainliest" that would be most appreciated
