[tex]QUADRATIC \: \: RESOLUTIONS \\ \\ \\
4 {x}^{2} \: + \: 40x \: + \: 84 \: = \: 0 \\ \\ 4 \: ( \: {x}^{2} + 10x + 21) \: = \: 0 \\ \\ 4 \: can \: never \: be \: equal \: to \: \: 0 \\ \\ {x}^{2} \: + \: 10x \: + \: 21 = \: 0 \\ \\ {x}^{2} + \: 7x \: + 3x \: + \: 21 = 0 \\ \\ x(x + 7) + 3(x + 7) = 0 \\ \\ ( \: x + 7 \: ) \: ( \: x + 3 \: ) \: = \: 0 \\ \\ On \: solving \: \: , \: \: \: We \: get \: \: \: \: \: - \\ \\ \\ x \: = \: - 7 \: \: \: or \: \: \: x = \: - 3 \: \: \: \: \: \: \: \: Ans.[/tex]
