Question

Resolva... - 16x² + 8x - 1 = 0

Answer 1

por bháskara tem-se:
x= -8 +-√8^2-4.(-16).(-1)/-32
x= -8 +- √64 -64/-32
x= -8/-32
x= 1/4

Answer 2

Resolução da questão pela fórmula de Bhaskara, veja:

$-16x^{2} + 8x-1=0\\\\\\\ x=-b~\pm~\dfrac{ \sqrt{b^{2}-4~.~a~.~c}}{2~.~a}}\\\\\\ x=-8~\pm~\dfrac{ \sqrt{8^{2}-4~.~(-16)~.~(-1)}}{2~.~(-16)}}\\\\\\ x=-8~\pm~\dfrac{ \sqrt{64-64}}{-32}}\\\\\\ x=-8~\pm~\dfrac{ \sqrt{0}}{-32}}\\\\\\ Quando~\Delta=0~x'=x''\\\\\\ x'=x''=\dfrac{-8}{-32}}\\\\\\ \large\boxed{\boxed{\boxed{\boxed{x'=x''=\dfrac{1}{4}.}}}}}}$

Espero que te ajude :).