Given a quadratic equation with the following form
[tex]ax^2+bx+c=0[/tex]
By the quadratic formula, the solutions are given by the following expression
[tex]x_{\pm}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]
In our problem we have the following equation
[tex]4x^2-7x+3=0[/tex]
Therefore, our coefficients are
[tex]\begin{gathered} a=4 \\ b=-7 \\ c=3 \end{gathered}[/tex]
Plugging those values into the quadratic formula, we have
[tex]x_{\pm}=\frac{-(-7)\pm\sqrt{(-7)^2-4(4)(3)}}{2(4)}[/tex]
Solving this equation, we have
[tex]\begin{gathered} x_{\operatorname{\pm}}=\frac{-(-7)\pm\sqrt{(-7)^2-4(4)(3)}}{2(4)} \\ =\frac{7\pm\sqrt{49-48}}{8} \\ =\frac{7\pm1}{8} \\ \implies\begin{cases}x_+={1} \\ x_-={\frac{3}{4}}=0.75\end{cases} \end{gathered}[/tex]
