Answer:
[tex](4x+1)(x-3)=0[/tex]
Step-by-step explanation:
we have
[tex]4x^2-11x-3=0[/tex]
Solve the quadratic equation
we know that
The formula to solve a quadratic equation of the form
[tex]ax^{2} +bx+c=0[/tex]
is equal to
[tex]x=\frac{-b\pm\sqrt{b^{2}-4ac}} {2a}[/tex]
in this problem we have
[tex]4x^2-11x-3=0[/tex]
so
[tex]a=4\\b=-11\\c=-3[/tex]
substitute in the formula
[tex]x=\frac{-(-11)\pm\sqrt{-11^{2}-4(4)(-3)}} {2(4)}[/tex]
[tex]x=\frac{11\pm\sqrt{169}} {8}[/tex]
[tex]x=\frac{11\pm13} {8}[/tex]
[tex]x=\frac{11+13} {8}=3[/tex]
[tex]x=\frac{11-13} {8}=-\frac{1}{4}[/tex]
therefore
Rewrite the expression
[tex]4x^2-11x-3=4(x+\frac{1}{4})(x-3)[/tex]
[tex]4x^2-11x-3=(4x+1)(x-3)[/tex]
[tex](4x+1)(x-3)=0[/tex]
