Question

How to solve 4n²-3n-7=0

Answer 1

$4n^2-3n-7=0\\\\\underbrace{(2n)^2-2\cdot2n\cdot\frac{3}{4}+\left(\frac{3}{4}\right)^2}_{(*)}-\left(\frac{3}{4}\right)^2=7\\\\(2n-\frac{3}{4})^2-\frac{9}{16}=7\\\\(2n-\frac{3}{4})^2=7+\frac{9}{16}\\\\(2n-\frac{3}{4})^2=\frac{112}{16}+\frac{9}{16}\\\\(2n-\frac{3}{4})^2=\frac{121}{16}\to2n-\frac{3}{4}=\sqrt\frac{121}{16}\ \vee\ 2n-\frac{3}{4}=-\sqrt\frac{121}{16}$

$2n=\frac{11}{4}+\frac{3}{4}\ \vee\ 2n=-\frac{11}{4}+\frac{3}{4}\\\\2n=\frac{14}{4}\ \vee\ 2n=-\frac{8}{4}\\\\2n=\frac{7}{2}\ \vee\ 2n=-2\\\\n=\frac{7}{4}\ \vee\ n=-1$



$(a-b)^2=a^2-2ab+b^2$

Answer 2

$4n^2-3n-7=0\\ \\ a=4, \ \ b=-3 \ \ c=-7\\ \\ \Delta = b^{2}-4ac = (-3)^{2}-4*4* (-7)= 9+112=121$

$x_{1}=\frac{-b-\sqrt{\Delta }}{2a} =\frac{ 3-\sqrt{ 121}}{2*4}=\frac{ 3-11}{8}= \frac{-8 }{ 8}=-1 \\ \\x_{2}=\frac{-b+\sqrt{\Delta }}{2a} =\frac{ 3+\sqrt{ 121}}{2*4}=\frac{ 3+11}{8}= \frac{14 }{ 8}= \frac{7}{4}=1\frac{3}{4}$