For a quadratic of the form [tex]f(x)=ax^2+bx+c[/tex], we have the quadratic formula
[tex]x=\dfrac{-b \pm \sqrt{b^2 -4ac} }{2a}[/tex],
where a is the coefficient (number before the variable) of the squared term, b is the coefficient of the linear term, and c is the constant term.
So, given [tex]2x^2-16x+27=0[/tex], we can get that [tex]a=2, \ b=-16[/tex], and [tex]c=27[/tex]. We substitute these numbers into the quadratic formula above.
[tex]x=\dfrac{-(-16) \pm \sqrt{(-16)^2 -4(2)(27)} }{2(2)} [/tex]
[tex]x=\dfrac{16 \pm \sqrt{(256 -216)} }{4} [/tex]
[tex]x=\dfrac{16 \pm \sqrt{40} }{4} [/tex]
[tex]x=\dfrac{16 \pm 2\sqrt{10} }{4} [/tex]
[tex]x=4+ \frac{\sqrt{10}}{2}, \ x=4- \frac{\sqrt{10}}{2}[/tex]
This is our final answer.
If you've never seen the quadratic formula, you can derive it by completing the square for the general form of a quadratic. Note that the [tex]\pm[/tex] symbol (read: plus or minus) represents the two possible distinct solutions, except for zero under the radical, which gives only one solution.
