Question

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$- 8 {x}^{2} + 40x = 0$
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Answer 1

Explicação passo-a-passo:

$-8x^2+40x=0\\\\=\frac{-40\pm \sqrt{40^2-4\left(-8\right)\cdot \:0}}{2\left(-8\right)}\\\\x_1=\frac{-40+40}{2\left(-8\right)}=\frac{-40+40}{-2\cdot \:8}=\frac{0}{-2\cdot \:8}=\frac{0}{-16}=0\\\\x_2=\frac{-40-40}{2\left(-8\right)}=\frac{-40-40}{-2\cdot \:8}=\frac{-80}{-16}=\frac{80}{16}=5\\\\x1=0, x2=5$