What are the zeros of the quadratic function f(x) = 2x2 – 10x – 3?

Answer: [tex]x=\frac{50+\sqrt{31}}{2},\frac{50-\sqrt{31}}{2}[/tex]  are zeroes of given quadratic equation.Step-by-step explanation:We have been a quadratic equation: [tex]2x^2-10x-3[/tex]We need to find the zeroes of quadratic equationWe have a formula to find zeroes of a quadratic equation:[tex]x=\frac{b^2\pm\sqrt{D}}{2a}\text{where}D=\sqrt{b^2-4ac}[/tex]General form of quadratic equation is [tex]ax^2+bx+c[/tex]On comparing general equation with b given equation we geta=2,b=-10,c=-3On substituting the values in formula we get[tex]D=\sqrt{(-10)^2-4(2)(-3)}[/tex][tex]\Rightarrow D=\sqrt{100+24}=\sqrt{124}[/tex]Now substituting D in  [tex]x=\frac{b^2\pm\sqrt{D}}{2a}[/tex] we get[tex]x=\frac{(-10)^2\pm\sqrt{124}}{2\cdot 2}[/tex][tex]x=\frac{100\pm\sqrt{124}}{4}[/tex][tex]x=\frac{100\pm2\sqrt{31}}{4}[/tex][tex]x=\frac{50\pm\sqrt{31}}{2}[/tex]Therefore, [tex]x=\frac{50+\sqrt{31}}{2},\frac{50-\sqrt{31}}{2}[/tex]