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# how do you find the zeros in

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how do you find the zeros in

y=-6x^2 + 42x -6

Guest Mar 29, 2017
#1
+93338
+2

a=-6, b=42, c=-6

Melody  Mar 29, 2017
#2
+13
+1

$$y= -6x^2+42x -6$$

First, factor out a -6

$$y = -6(x^2 - 7x + 1)$$

Then, set the value in the parentheses equal to 0

$$x^2-7x+1=0$$

Now, use the quadratic formula to solve

$$y = {(7) \pm \sqrt{(7)^2-4(1)(1)} \over 2(1)}$$

$$y = {7 \pm 3 \sqrt5 \over 2}$$

$$y \approx 6.854, 0.146$$

Your zeros are approximately 6.854, and 0.146

austinruffino  Mar 29, 2017