How do you find the zeros, real and imaginary, of $y = - 2 x^{2} - 9 x + 5$ $y=-2x^2-9x+5$ using the quadratic formula?

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Answer 1 · 2018-04-22T11:04:29

Zeros are real and $x = 0.5, x = - 5$ $x=0.5, x= -5$

$y = - 2 x^{2} - 9 x + 5$ $y= -2 x^2-9 x +5$

Comparing with standard quadratic equation $a x^{2} + b x + c = 0$ $ax^2+bx+c=0$

$a = - 2, b = - 9, c = 5$ $a=-2 ,b=-9 ,c=5$ . Discriminant $D = b^{2} - 4 a c$ $D= b^2-4 a c$ or

$D = 81 + 40 = 121$ $D=81+40 =121$ , discriminant is positive, we get two real

solutions. Quadratic formula: $x = \frac{- b \pm \sqrt{D}}{2 a}$ $x= (-b+-sqrtD)/(2a)$ or

$x= (9+-sqrt 121)/(-4) = (9 +- 11)/-4 :. x= 20/-4= -5$ or

$x=(-2)/-4 = 0.5 :.$ Zeros are real and $x=0.5, x= -5$ [Ans]

How do you find the zeros, real and imaginary, of y=-2x^2-9x+5y=−2x2−9x+5y=-2x^2-9x+5 using the quadratic formula?