Question

$- {y}^{2} + 13y - 40 = 0$
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Answer 1

Resposta:

Explicação passo-a-passo:

$y=\frac{-13+-\sqrt{13^2-4.(-1).(-40)} }{2.(-1)} =\frac{-13+-\sqrt{169-160} }{-2} =\frac{-13+-\sqrt{9} }{-2}= \frac{-13+-3}{-2} \\ \\ y_{1}=\frac{-13+9}{-2}=\frac{-4}{-2}=2\\ \\ y_{2}=\frac{-13-9}{-2}=\frac{-22}{-2}=11$