How do you find the roots, real and imaginary, of $y=10x^2 + 9x-1$ using the quadratic formula?

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Answer 1 · 2017-01-15T03:08:41

$x_(1,2)=(-B+-sqrt(B^2-4AC))/(2A)$

$x_(1,2)=(-9+-sqrt((-9)^2-4*10*(-1)))/(2*10)$

$x_(1,2)=(-9+-sqrt(81+40))/(20)$

$x_(1,2)=-9/20+-1/20sqrt121=-9/20+-11/20$

$->x_1=-9/20+11/20=2/20=1/10$

$->x_2=-9/20-11/20=-20/20=-1$

How do you find the roots, real and imaginary, of y=10x^2 + 9x-1 y=10x^2 + 9x-1 using the quadratic formula?