Answer:
[tex]x=\frac{5+\sqrt{19}}{2}\text{ and } x=\frac{5-\sqrt{19}}{2}[/tex]
Step-by-step explanation:
If we have the standard form [tex]ax^2+bx+c[/tex], then we can use the quadratic formula:
[tex]x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]
First, let's identify our coefficients. We have [tex]2x^2-10x+3[/tex].
This can be rewritten as [tex](2)x^2+(-10)x+(3)[/tex].
Therefore, a=2, b=-10, and c=3.
Substitute these values into the quadratic formula. This yields:
[tex]x=\frac{-(-10)\pm\sqrt{(-10)^2-4(2)(3)}}{2(2)}[/tex]
From here, simplify. Evaluate the expression under the square root:
[tex]x=\frac{10\pm\sqrt{100-24}}{4}[/tex]
Evaluate:
[tex]x=\frac{10\pm\sqrt{76}}{4}[/tex]
Note that:
[tex]\sqrt{76}=\sqrt{4\cdot 19}=\sqrt{4}\cdot\sqrt{19}=2\sqrt{19}[/tex]
Therefore:
[tex]x=\frac{10\pm2\sqrt{19}}{4}[/tex]
We can factor out a 2 from both the numerator and the denominator:
[tex]x=\frac{2(5\pm\sqrt{19})}{2(2)}[/tex]
Simplify:
[tex]x=\frac{5\pm\sqrt{19}}{2}[/tex]
Therefore, our roots are:
[tex]x=\frac{5+\sqrt{19}}{2}\text{ and } x=\frac{5-\sqrt{19}}{2}[/tex]
