Question

f ) y (y + 3) = 40

me ajudem por favor! ​

Answer 1

Olá!

y · (y + 3) = 40

y² + 3y = 40

y² + 3y - 40  =  0      →     Bháskara

$\dfrac{-b\pm\sqrt{b^{2}-4\cdot a\cdot c} }{2\cdot a}~~~\to~~~\dfrac{-3\pm\sqrt{3^{2}-4\cdot 1\cdot (-40)} }{2\cdot 1}~~~\to ~~~\dfrac{-3\pm\sqrt{169} }{2} ~~~\to ~~~\dfrac{-3\pm13}{2} \\ \\ \\ \\ \boxed{y'=5~~~e~~~y''=-8}$

:)