ANSWER:
The zeros of the function are 5.35 and 0.65
STEP-BY-STEP EXPLANATION:
We have the following function:
[tex]f(x)=2x^2-12x+7[/tex]
We equal 0 and use the general formula of quadratic equations to calculate the zeros of the function, just like this:
[tex]\begin{gathered} x_{1,\:2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a} \\ \\ a=2,\:b=-12,\:c=7 \\ \\ x_{1,\:2}=\frac{-\left(-12\right)\pm \sqrt{\left(-12\right)^2-4\cdot \:2\cdot \:7}}{2\cdot \:2} \\ \\ x_{1,2}=\frac{12\pm\sqrt{88}}{4} \\ \\ x_1=\:\frac{12+\sqrt{88}}{4}\:=5.35 \\ \\ x_2=\:\frac{12-\sqrt{88}}{4}\:=0.65 \end{gathered}[/tex]
Therefore, the zeros of the function are 5.35 and 0.65
