Question Number 90316 by M±th+et£s last updated on 22/Apr/20
$$\begin{vmatrix}{{x}}&{\mathrm{7}}\\{\mathrm{9}}&{\mathrm{8}−{x}}\end{vmatrix}=\begin{vmatrix}{\mathrm{7}}&{\mathrm{0}}&{−\mathrm{3}}\\{−\mathrm{5}}&{{x}}&{−\mathrm{6}}\\{−\mathrm{3}}&{−\mathrm{5}}&{{x}−\mathrm{9}}\end{vmatrix} \\ $$$$ \\ $$
Commented by john santu last updated on 23/Apr/20
$$\mathrm{8}{x}−{x}^{\mathrm{2}} −\mathrm{63}\:=\:\mathrm{7}\left({x}^{\mathrm{2}} −\mathrm{9}{x}−\mathrm{30}\right)− \\ $$$$\mathrm{3}\left(\mathrm{25}+\mathrm{3}{x}\right) \\ $$$$−{x}^{\mathrm{2}} +\mathrm{8}{x}−\mathrm{63}=\mathrm{7}{x}^{\mathrm{2}} −\mathrm{63}{x}−\mathrm{210}−\mathrm{75}−\mathrm{9}{x} \\ $$$$−{x}^{\mathrm{2}} +\mathrm{8}{x}−\mathrm{63}=\mathrm{7}{x}^{\mathrm{2}} −\mathrm{72}{x}−\mathrm{285} \\ $$$$\mathrm{8}{x}^{\mathrm{2}} −\mathrm{80}{x}−\mathrm{222}=\mathrm{0} \\ $$$$\mathrm{4}{x}^{\mathrm{2}} −\mathrm{40}{x}−\mathrm{111}=\mathrm{0} \\ $$$${x}\:=\:\frac{\mathrm{40}\:\pm\:\sqrt{\mathrm{40}^{\mathrm{2}} +\mathrm{444}}}{\mathrm{8}} \\ $$
Commented by M±th+et£s last updated on 23/Apr/20
$${sir}\:{i}\:{think}\:{its}\:\mathrm{8}{x}^{\mathrm{2}} −\mathrm{80}{x}−\mathrm{222}\:{and} \\ $$$${thank}\:{you}\:{now}\:{ican}\:{do}\:{it}\:{with}\:{my}\:{self} \\ $$
Answered by M±th+et£s last updated on 23/Apr/20
Commented by john santu last updated on 23/Apr/20
$${yes}.\:{good} \\ $$