Let x be the number.
We know that we squared ir and doubled it, this means:
[tex]2x^2[/tex]This have to be equal to four more than the product of seven and the number:
[tex]2x^2=7x+4[/tex]this is the same as:
[tex]2x^2-7x-4=0[/tex]this can be solved for the general formula:
[tex]x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}[/tex]in our case a=2, b=-7 and c=-4; then:
[tex]\begin{gathered} x=\frac{-(-7)\pm\sqrt[]{(-7)^2-4(2)(-4)}}{2(2)} \\ x=\frac{7\pm\sqrt[]{49+32}}{4} \\ x=\frac{7\pm\sqrt[]{81}}{4} \\ x=\frac{7\pm9}{4} \end{gathered}[/tex]hence:
[tex]\begin{gathered} x=\frac{7+9}{4}=\frac{16}{4}=4 \\ or \\ x=\frac{7-9}{4}=-\frac{2}{4}=-\frac{1}{2} \end{gathered}[/tex]Since we are looking for a whole number, then the number is 4.
