How do you find the roots, real and imaginary, of $y= -3x^2-16x +8$ using the quadratic formula?

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Answer 1 · 2017-10-22T20:40:04

Roots are $**5.7936, -0.4603**$

Standard form of equation is $ax^2 + bx + c = 0$
Roots are $= ((-b) +- sqrt(b^2 -(4ac)))/(2a)$

Given equation is $-3x^2 -16x + 8 = 0$
$a = -3, b = -16, c = 8$

$x =( -(-16) +- sqrt((-16)^2 - (4*(-3)8)))/(2*(-3))$

$x = (16 +- sqrt(256 + 96))/6$

$x = (16 +- sqrt352)/6$

$x = 5.7936, -0.4603$

How do you find the roots, real and imaginary, of y= -3x^2-16x +8 y= -3x^2-16x +8 using the quadratic formula?