bonjourr! j'ai ce petit probleme x²-(x+3)(4x+5)=9 resoudre ce probleme MERCI D'AVANCE

bonjour

x² - ( x + 3 ) ( 4 x + 5 ) = 9

x² - ( 4 x² + 5 x + 12 x + 15 ) = 9

x² -  4 x² - 5 x - 12 x - 15 =  9

- 3 x² - 17 x - 15 - 9 = 0

- 3 x² - 17 x - 24 = 0

Δ = ( - 17 )² - 4 ( - 3 * - 24 ) = 289 - 288 = 1

x 1 = ( 17 - 1 ) / - 6 = 16 / - 6 = - 8/3

x 2 = ( 17 + 1 ) / - 6 = - 18 /6 = - 3

Bjr

x²-(x+3)(4x+5) = 9

x²-(4x²+5x+12x+15) = 9

x²-(4x²+17x+15) = 9

x²-4x²-17x-15 = 9

-3x²-17x-15 = 9

-3x²-17x-15-9 = 0

-3x²-17x-24 = 0

3x²+17x+24 = 0

3x²+9x+8x+24 = 0

3x×(x+3)+8x+24 = 0

3x×(x+3)+8(x+3) = 0

(x+3)(3x+8) = 0

x+3 = 0 / 3x+8 = 0

x=-3 / 3x+8 = 0

x=-3 / x=-8/3

x1 = -3 / x2 = -8/3

Voilà !